GATE Overflow - Recent questions and answers in Set Theory & Algebra
http://gateoverflow.in/qa/mathematics/discrete-mathematics/set-theory-%26-algebra
Powered by Question2AnswerAnswered: Problem related to irreflexive (Relation)
http://gateoverflow.in/134951/problem-related-to-irreflexive-relation?show=134973#a134973
If any website contains any link then the relation holds for that page (as the link is common to itself) making the relation irreflexive. Only if we restrict the domain of webpages to those without any links, the relation will be irreflexive.Set Theory & Algebrahttp://gateoverflow.in/134951/problem-related-to-irreflexive-relation?show=134973#a134973Tue, 27 Jun 2017 15:32:19 +0000Answered: GATE2005-IT-33
http://gateoverflow.in/3779/gate2005-it-33?show=134878#a134878
<p><strong>Lets take an example..</strong></p>
<p>Suppose A= {1,2,3} . here n=3</p>
<p>Now P(A)= {∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}</p>
<p><strong>now C will contain ∅ (empty set) and ,{1,2,3} (set itself) as ∅ is the subset of every set. And every other subset is the subset of {1,2,3}.</strong></p>
<p>not taking subsets of cardinality 1. </p>
<p>We can take any 1 of {1},{2},{3} as none of the set is subset of the other.</p>
<p>Lets take {2}</p>
<p>Now taking the sets of cardinality 2- {1,2},{1,3},{2,3} .</p>
<p>{2}⊂ {1,2} and {2,3} but we can't take both as none of the 2 is subset of the other.</p>
<p>so lets take {2,3}</p>
<p>so C = {∅,{2},{2,3},{1,2,3}} </p>
<p>So if we observe carefully . We can see that we can select only 1 set from the subsets of each cardinality 1 to n . </p>
<p>i.e total n subsets + ∅ = n+1 subsets of A can be there in C</p>
<p>So even though we can have different combinations of subsets in C but maximum cardinality of C will be <strong>n+1</strong> only.</p>
<p><strong>So B is the answer. </strong></p>Set Theory & Algebrahttp://gateoverflow.in/3779/gate2005-it-33?show=134878#a134878Tue, 27 Jun 2017 02:20:12 +0000Answered: TIFR2014-B-18
http://gateoverflow.in/27351/tifr2014-b-18?show=134876#a134876
The function $g_z(Y)$ is defined as $[k] \to \{0,1\}$ where $[k]$ is the set of positive integers till $k$. That is, given a triplet $(z_1,z_2, z_3)$, $Y$ can take any value from $1$ to $k$. If $Y$ happen to be any of $z_1, z_2, z_3$, $g_z(Y) = 0$ due to the definition of $f$ and $g_z$. Now even for different $z$, $g_z$ may be the same. Otherwise, the answer would have been how many ways we can form a triplet $z$ - which gives $k^3$ and for each $z$ we get a function $g_z$.<br />
<br />
For all unique combinations of $z_1, z_2, z_3$ are unique, we are guaranteed that we get a distinct function $g_z$. This is clear from the definition of $g_z$. For example, suppose $k=4$. The triplets are<br />
<br />
$(1,2,3)$<br />
$(1,2,4)$<br />
$(1,3,4)$<br />
$(2,3,4)$<br />
For the triplet $(1,2,3)$, $Y$ can be made in $4$ ways as $(1,2,3,1), (1,2,3,2), (1,2,3,3)$ and $(1,2,3,1)$. Now, as per definition of $g_z$, we get $g_{(1,2,3)} = \{\{1 \to 0\},\{ 2 \to 0\},\{ 3 \to 0\},\{ 4 \to 1\}\}$.<br />
<br />
Similarly, for the next three triplets, $g_z$ are different as in second only $3$ maps to $1$, in third one only $2$ maps to $1$ and in fourth one only $1$ maps to $1$.<br />
<br />
So, in general, for any given $k$, we have ${}^kC_3$ ways of forming distinct triplets and each of them guarantees a unique function $g_z$ where exactly $k-3$ elements map to $1$ and $3$ elements map to $0$. Now, if any of the elements in the triplet are same, then the function becomes $\{\{1 \to 0\}, \{2 \to 0 \}, \dots, \{k \to 0\}\}$, (all $k$ elements mapping to $0$) and this remains the same for any triplet. So, total number of possible functions are<br />
<br />
$${}^kC_3 + 1$$Set Theory & Algebrahttp://gateoverflow.in/27351/tifr2014-b-18?show=134876#a134876Mon, 26 Jun 2017 22:49:42 +0000Answered: GATE2014-2-50
http://gateoverflow.in/2016/gate2014-2-50?show=134839#a134839
<p><strong>Symmetric difference (S.D)- </strong>suppose A and B are 2 sets</p>
<p>then symmetric difference of A and B is (A-B)U(B-A) which is equal to (AUB)-(A⋂B)</p>
<p><strong>Now Consider a smaller set.</strong></p>
<p>Suppose S= {1,2,3,4}</p>
<p>now the given 2 statements are about smallest and largest subset.</p>
<p>So considering set S and ∅ (empty set) will be helpful.</p>
<p>First take U = {1,2,3,4}</p>
<p>V = {1,2} (we can take any set other than ∅ and S)</p>
<p>S.D = {3,4} (just exclude the elements which are common in the 2 sets)</p>
<p>minimum element of S.D is 3 which is in U. </p>
<p>and if we observe carefully minimum element will always be in U .Whatever the V is.</p>
<p>So acc. to the question {1,2,3,4} is smaller than any other subset of S. <strong>S2 is true.</strong></p>
<p>Now consider </p>
<p>U= {∅}</p>
<p>V= {1,2} (we can take any subset of S)</p>
<p>S.D = {1,2}</p>
<p>Symmetric difference will be always equal to V . So minimum element of S.D will always exist in V when U is {∅}.</p>
<p>So acc. to the que, {∅} is the greater than any other subset of S. <strong>S1 is also true.</strong></p>
<p>This is true even when S= {1,2,3......2014}.</p>
<p><strong>So answer is A. Both S1 and S2 are true</strong></p>Set Theory & Algebrahttp://gateoverflow.in/2016/gate2014-2-50?show=134839#a134839Mon, 26 Jun 2017 13:40:52 +0000Answered: Hasse diagram
http://gateoverflow.in/134666/hasse-diagram?show=134670#a134670
$D_n$ is boolean algebra if and only if $n$ is a product of distinct primes. That is all primes factors appear only once in the prime factorization of $n$.<br />
<br />
$66 = 2 * 3 * 11 \;\;\;\;\; $ Therefore $D_{66}$ is boolean algebra.Set Theory & Algebrahttp://gateoverflow.in/134666/hasse-diagram?show=134670#a134670Sun, 25 Jun 2017 04:50:30 +0000Answered: GATE2003-38
http://gateoverflow.in/929/gate2003-38?show=134455#a134455
<p> </p>
<p>We want<strong> operand 1 + operand 2=c ,</strong></p>
<p>so the pairs can be:
<br>
(a, c)</p>
<p>(b, c)</p>
<p>(c, b)</p>
<p><strong>now note that identity element of ({a,b,c},*) is 'a'</strong></p>
<p>so (a∗x)+(a∗y)=c will results in the above 3 pair only.</p>
<p>so no need to check equation 1.</p>
<p>just see if these pairs satisfy (b∗x)+(c∗y)=c.</p>
<p>.(b, c) and (c, b) satisfy equation 2 also so there are 2 solutions.</p>
<p><strong>C is the answer.</strong></p>Set Theory & Algebrahttp://gateoverflow.in/929/gate2003-38?show=134455#a134455Fri, 23 Jun 2017 08:27:11 +0000Answered: TESTBOOK TEST
http://gateoverflow.in/134352/testbook-test?show=134362#a134362
<p style="text-align:center"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=13796581093940849814"></p>
<p>$60 + 20 - x + a = 100$</p>
<p>$ 60 + 40 -x + c=100$</p>
<p>$ 20 + 40 - x + b=100$</p>
<p>$a + b + c = 210$</p>
<p>$x=50$</p>Set Theory & Algebrahttp://gateoverflow.in/134352/testbook-test?show=134362#a134362Thu, 22 Jun 2017 16:46:27 +0000Answered: discrete mathematics , group theory
http://gateoverflow.in/133661/discrete-mathematics-group-theory?show=134185#a134185
for a group to be abelian following properties must be satisfied<br />
<br />
(1)closure<br />
<br />
(2)associ.<br />
<br />
(3)identity<br />
<br />
(4)inverse<br />
<br />
(5)commu.Set Theory & Algebrahttp://gateoverflow.in/133661/discrete-mathematics-group-theory?show=134185#a134185Wed, 21 Jun 2017 13:29:31 +0000Answered: ISI2017 MMA
http://gateoverflow.in/132738/isi2017-mma?show=133895#a133895
good question.<br />
<br />
yes u r correct . because we do not know the equation of function hence we can not predict at what point it is discontinuous. but we can definitely say that open interval (-infinity to +infinity) will include all possible number ,hence whatever will the function be , at some point it may violate the given condition . therefore option D is correct oneSet Theory & Algebrahttp://gateoverflow.in/132738/isi2017-mma?show=133895#a133895Mon, 19 Jun 2017 15:00:40 +0000Answered: TIFR2017-A-11
http://gateoverflow.in/95289/tifr2017-a-11?show=133837#a133837
$f:A\rightarrow B$ is injective if and only if, given any functions $g,h:B\rightarrow A$ whenever $f\circ g=f\circ h,$ $f\circ g=f\circ h,then\ g=h$.<br />
<br />
Refer to properties of Injective functions: <a href="https://en.wikipedia.org/wiki/Injective_function" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Injective_function</a><br />
<br />
Let us prove $(\forall g,h:f(g(x))=f(h(x))\rightarrow g(x)=h(x))\rightarrow f\ is\ one-to-one$ is true.<br />
<br />
This is equivalent to, $f\ is\ not\ one-to-one\rightarrow (\exists g,h:f(g(x))=f(h(x))\wedge g(x)\neq h(x))$<br />
<br />
Let us assume LHS is true, i.e. $f\ is\ not\ one-to-one$.<br />
<br />
Then there exists some $c,d\in A$ such that,<br />
<br />
$f(c)=f(d)=a,$ where $a$ is an arbitrary element which belongs to B<br />
<br />
Let g and h be some functions out of all possible functions from B to A such that $g\neq h$,<br />
<br />
i.e. $g(x)=c\ and\ h(x)=d\ \exists c,d\in A\ and\ \exists x\in B$<br />
<br />
$\therefore f(g(x))=f(c)=a\ and\ f(h(x))=f(d)=a$ and $g(x)\neq h(x)$, i.e. RHS is also true.<br />
<br />
Thus, whenever $\forall g,h\ f\circ g=f\circ h\rightarrow g=h$ is true, $f \ is\ one-to-one$.<br />
<br />
<br />
<br />
Domain of $f$ need not be finite. Let $f:A\rightarrow B$ be identity function and A and B be infinite sets. Assume that $f\circ g=f\circ h$ is true,<br />
<br />
then $f(g(x))=f(h(x))\rightarrow g(x)=h(x))$ will be true $\forall g,h$ since $f$ is an identity function. So, even if domain of f is not finite, the condition holds true.Set Theory & Algebrahttp://gateoverflow.in/95289/tifr2017-a-11?show=133837#a133837Mon, 19 Jun 2017 10:24:30 +0000Answered: Set theory
http://gateoverflow.in/133317/set-theory?show=133321#a133321
{a} ∈ A means {a} must be present as element in set A.<br />
<br />
whereas,a ∈ A means only a must be present as element in A<br />
<br />
Example:let A={ {a},b,c,{b,c}} then {a} ∈ A is correct but a ∈ A is NOT.<br />
<br />
and let B={a,b,{c,d}} then a ∈ B is correct but {a} ∈ B is not.<br />
<br />
HOPE IT WILL BE CLEAR NOW :)Set Theory & Algebrahttp://gateoverflow.in/133317/set-theory?show=133321#a133321Fri, 16 Jun 2017 05:33:14 +0000Answered: GATE1988-13ii
http://gateoverflow.in/94634/gate1988-13ii?show=132916#a132916
let set s={1,2,3,4} <br />
<br />
now see mapping from s to s<br />
<br />
for f to be onto every element of codomain must be mapped by every element in domain.<br />
<br />
since cardinality is same for both domain and codomain. we can not have mapping like f(1)=1 & f(2)=1 if it happened then at least one element remain umapped in codomain,which resultant f not to be onto but it is given that f is onto.so every element in codomain have exactly one element in domain.so one of mapping be like f(1)=2, f(2)=3,f(3)=4,f(4)=1 which certainly prove that f is an one one function also.<br />
<br />
NOTE:if s is infinite then this result may not always be true.Set Theory & Algebrahttp://gateoverflow.in/94634/gate1988-13ii?show=132916#a132916Tue, 13 Jun 2017 15:13:59 +0000Answered: Series Summation
http://gateoverflow.in/132690/series-summation?show=132695#a132695
Take your final term as $$\frac{1}{(n-2)(n-1)(n)(n+1)}$$<br />
<br />
If we try to write it in another form, we have $$\frac{1}{3}\left ( \frac{1}{(n-2)(n-1)(n)}-\frac{1}{(n-1)(n)(n+1)} \right )$$<br />
<br />
OR<br />
<br />
$$\frac{1}{3}\left ( X_{n}-X_{n+1}\right )$$<br />
<br />
Can you proceed from here by taking summation ?Set Theory & Algebrahttp://gateoverflow.in/132690/series-summation?show=132695#a132695Sun, 11 Jun 2017 17:28:08 +0000Answered: Kenneth Rosen Edition7 Ch-7 Ex-1 QueNo-5
http://gateoverflow.in/131099/kenneth-rosen-edition7-ch-7-ex-1-queno-5?show=131101#a131101
<p><strong>(a)</strong></p>
<p><em><strong>1) reflexive :</strong></em>Everyone who has visited web page a also has visited web page a</p>
<p><em><strong>2) </strong></em><em><strong>symmetric: </strong></em>There are web pages a and b such that the set people who have visited web page a also visited web page b but there may be other people visiting web page b may not have visited web page a (e.g., depends on links between web page a and web page b). so not symmetric
<br>
<em><strong>3) anti symmetric:</strong></em> It is conceivable that there are the same set of visitors for web page a and b. so not anti symmetric.
<br>
<em><strong>4) transitive</strong></em> :If everyone who has visited a has visited b and everyone who has visited b has visited c implies that everyone who has visited a has visited c. so transitive</p>Set Theory & Algebrahttp://gateoverflow.in/131099/kenneth-rosen-edition7-ch-7-ex-1-queno-5?show=131101#a131101Sat, 27 May 2017 16:36:48 +0000Answered: $a_n = 4^n + 6^n$
http://gateoverflow.in/130200/%24a_n-4-n-6-n%24?show=130210#a130210
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=14247151916419105395"></p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=12355190060415847842"></p>
<p>correct me if i m wrong....</p>Set Theory & Algebrahttp://gateoverflow.in/130200/%24a_n-4-n-6-n%24?show=130210#a130210Sat, 20 May 2017 05:17:01 +0000Answered: The gatebook
http://gateoverflow.in/129333/the-gatebook?show=130199#a130199
For a Poset to be a lattice:<br />
<br />
i. Every pair of elements should have LUB (Least upper bound).<br />
<br />
ii. Every pair of elements should have GLB (Greatest lower bound).<br />
<br />
As we know, a Poset in which every element is related, is called Total ordered set. The Hasse diagram forms a linear chain. In such a chain we can easily find LUB and GLB for every pair of elements. <br />
<br />
Hence Total ordered set is a lattice.Set Theory & Algebrahttp://gateoverflow.in/129333/the-gatebook?show=130199#a130199Fri, 19 May 2017 19:39:03 +0000Answered: ISRO2017-9
http://gateoverflow.in/128555/isro2017-9?show=129557#a129557
option B is My AnswerSet Theory & Algebrahttp://gateoverflow.in/128555/isro2017-9?show=129557#a129557Sun, 14 May 2017 01:52:22 +0000Answered: GATE2000-6
http://gateoverflow.in/677/gate2000-6?show=129356#a129356
<p>$S = {1,2,3,4,5,6.....n}$</p>
<p>Let us assume any two subset $S_1$ and $S_2$. We can simply assume $n(S_1 \cap S_2) =0$ to consider the disconnected sets if we want. </p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=17761882529809928727"></p>
<p> </p>
<p>Now there are three cases in which $(S_1 \backslash S_2) \cup (S_2 \backslash S_1) \;\; Or, \;\; (S_1 \oplus S_2) $ has only $2$ element.</p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=8574564280166559846"></p>
<p> </p>
<ol>
<li>Both green shaded area has one element each and in this case sizes of $S_1$ and $S_2$ are same.</li>
<li>The green area of $S_1$ contains $2$ element and the green area of $S_2$ contains none. In this case size of $S_1$ is $2$ more than that of $S_2$.</li>
<li>The green area of $S_2$ contains $2$ element and the green area of $S_1$ contains none. In this case size of $S_2$ is $2$ more than that of $S_1$.</li>
</ol>
<p> </p>
<p>So, if we are only interested in a <strong>particular </strong><span class="marker">set vertex</span> corresponding to set $S_1$ of size $= m$, then $S_1$ is connected to <strong>three types</strong> of set vertices as shown below. We will use the words "<strong>set</strong>" and "<strong>vertices</strong>" synonymously. </p>
<p> <img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11969504495141772952"></p>
<p>In this above image, we have considered $m \geq 2$. The cases for $m = 1 \text{ and } m = 0$ will be <strong>discussed later</strong>.</p>
<p>Now, what we need to find is the <strong>no of set vertices</strong> in each of the <strong>above</strong> three types and sum them up to get the <strong>degree</strong> of the vertex corresponding to the set $S_1$.</p>
<p>For simplicity let us assume $S = \{1,2,3,4,5,6,7\}$ and set $S_1 = \{1,2,3,4\}$. Our interest will be to find $S_2$ such that vertices corresponding to $S_1$ and $S_2$ are connected.</p>
<ol>
<li><strong>CASE 1</strong> : If we try to find another set $S_2$ having $4$ elements and satisfying constraint $n(S_1 \oplus S_2) = 2$, then we will see that no of such set $S_2$ is $4 \cdot (7 - 4)$. Or in general if $S_1$ is an $m$ element set then no of such $S_2$ sets with constraint $n(S_1 \oplus S_2) = 2$ will be equal to $m\cdot (n-m)$.</li>
<li><strong>CASE 2 </strong>: $S_1$ contains $4$ element and If we try to find $S_2$ where $S_2$ contains $2$ elements and satisfying constraint $n(S_1 \oplus S_2) = 2$, then no of such $S_2$ will be $4C2$ or in general, for $m$ element set $S_1$, we have $mC2$ no of $S_2$ type sets all with $(m-2)$ size.</li>
<li><strong>CASE 3</strong>: $S_1$ contains $4$ element and If we try to find $S_2$ where $S_2$ contains $6$ element and satisfying constraint $n(S_1 \oplus S_2) = 2$, then no of such $S_2$ sets will be $3C2$ or $(7-4)C2$. In general, with $S_1$ being $m$ element set, then $(n-m)C2$ no of $S_2$ sets will be possible.</li>
</ol>
<p> </p>
<p>Therefore, summing all three cases :</p>
<p>Degree of vertex $S_1$ ( assuming general case of $n(S_1) = m$ )</p>
<p>$\begin{align*}
<br>
&=m\cdot (n-m) + \binom{m}{2} + \binom{n-m}{2} \\
<br>
&=m\cdot n - m^2 + \frac{m^2}{2} - \frac{m}{2} + \frac{(n-m)\cdot (n-m-1)}{2} \\
<br>
&=m\cdot n - m^2 + \frac{m^2}{2} - \frac{m}{2} + \frac{n\cdot (n-1)}{2} \\
<br>
&\qquad - \frac{n \cdot m}{2} - \frac{n \cdot m}{2} + \frac{m^2}{2} + \frac{m}{2} \\
<br>
&=\frac{n\cdot (n-1)}{2} \\
<br>
&=\binom{n}{2} \\
<br>
\end{align*}$</p>
<p>This result is independent of $m$ for $m \geq 2$ and $m \leq n$.</p>
<p>For $m = 0$ and $m = 1$ also we can show that degree of $0$ and $1$ size set vertices is nothing but $nC2$ only. (fairly straight forward cases).</p>
<p>So we can conclude that every vertex has the same degree and the degree is $nC2$.</p>
<p> </p>
<hr>
<p> </p>
<p>Now we can guess one thing by looking at the following image:</p>
<p> </p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11969504495141772952"></p>
<p> </p>
<p>i.e.for $m \geq 2$ if $m$ is even the $S_1$ is connected to only even cardinality type of sets (<strong>at least one</strong>) or if $m$ is odd then $S_1$ is connected to only odd cardinality type of sets (<strong>at least one</strong>). By this, we can almost say that there are two connected components in the graph.</p>
<p>But there is little more argument before we can proceed and have a valid proof.</p>
<p>if $m = 0$ then $S_1 = \phi$, Then $S_1$ will be connected to all $m = 2$ type of sets or $2$ cardinality sets.</p>
<p>if $m = 1$ then $S_1$ will be one of all $1$ element sets, Then $S_1$ will be connected to all other $1$ cardinality sets and at least one $3$ cardinality set.</p>
<p>We can argue that, one $m$ (even) cardinality set is at least connected to one $(m-2)$ cardinality set. That particular $(m-2)$ cardinality set is at least connected to one $(m-4)$ cardinality set and so on <span class="marker">till $\phi$ set vertex</span>. There for all even cardinality sets are connected to $\phi$ directly or indirectly. </p>
<p>A similar argument holds for odd cardinality set vertices till we reach some $1$ cardinality set. <span class="marker">Moreover</span><span class="marker"> all $1$ cardinality sets are connected</span>. </p>
<p>Therefore we have a situation now that all even cardinality sets form one connected component and all odd cardinality set form another component.</p>
<p>For example : $n = 4$ :</p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=7495116809386705926"></p>Set Theory & Algebrahttp://gateoverflow.in/677/gate2000-6?show=129356#a129356Fri, 12 May 2017 08:09:13 +0000Answered: type of relation
http://gateoverflow.in/109601/type-of-relation?show=129264#a129264
x R x -x is related to x by 0 min since relation is at most 20 min apart <br />
<br />
xRy == yRx this holds as at most x and y are seperated by 20 min .<br />
<br />
xRy,yRz but x is not related to z , there may be a time of 40 min seperation at mostSet Theory & Algebrahttp://gateoverflow.in/109601/type-of-relation?show=129264#a129264Thu, 11 May 2017 09:39:59 +0000Answered: keneth r rosen discrete mathematics
http://gateoverflow.in/129180/keneth-r-rosen-discrete-mathematics?show=129245#a129245
<p> A relation R on a set S is called a partial. ordering if it is reflexive, antisymmetric and transitive</p>
<p>1) reflexive:- binary <strong>relation</strong> R over a set X is reflexive if ∀x ∈ X : x R x</p>
<p> x R x ie x=2x which is not satisfy</p>
<p>so not reflexive</p>
<p>2) anti symmetric :-R is <strong>anti-symmetric</strong> precisely if for all x and y in X. if x R y and y R x then x =y</p>
<p>x R y: x=2y</p>
<p>and y R x: y=2x</p>
<p>it does not mean that x=y </p>
<p>so not antisymmetric</p>
<p>3) transtive:- if x R y and y R x then x R Z</p>
<p>x R y:x=2y</p>
<p>y R z: y=2z</p>
<p>then from above x=4z (but not x=2z for the condition x R z)</p>
<p>so not transitive</p>
<p>so not POSET</p>Set Theory & Algebrahttp://gateoverflow.in/129180/keneth-r-rosen-discrete-mathematics?show=129245#a129245Thu, 11 May 2017 07:01:33 +0000Answered: GATE1994-1.4, ISRO2017-2
http://gateoverflow.in/2441/gate1994-1-4-isro2017-2?show=128797#a128797
<p>If <strong>A</strong> and<strong> B</strong> are independent events then:</p>
<p><strong>A$\cap$B = $\Phi$</strong> and <strong>A$\cup$B = A+B</strong></p>
<p>If <strong>A</strong> and<strong> B</strong> are non distinct events then:</p>
<p><strong>A$\cap$B != $\Phi$</strong> and <strong>A$\cup$B = A+B-A$\cap$B</strong> <em><strong>(principle of inclusion-exclusion)</strong></em></p>
<p>therefore <strong>A$\cup$B <= A + B</strong></p>
<p>so<strong> D</strong> is the answer...</p>Probabilityhttp://gateoverflow.in/2441/gate1994-1-4-isro2017-2?show=128797#a128797Sun, 07 May 2017 23:16:44 +0000Answered: PGEE 2017
http://gateoverflow.in/127522/pgee-2017?show=128409#a128409
Although I have not given this Exam, Still trying.<br />
<br />
<br />
<br />
Statement 1- False. <br />
<br />
Reason- Number of one one function from set A containing n elements to set B containing m elements is mpn.<br />
<br />
ATQ- npn= n!. <br />
<br />
Statement 2- True. <br />
<br />
Suppose domain and codomain is set of all integers (Z). Then <br />
<br />
ATQ- Z^Z -----> Z^Z is same / equivalent to Z ------> Z. [ as integers are closed under multiplication ]. <br />
<br />
It will be bijective. <br />
<br />
<br />
<br />
Statement 3- False. <br />
<br />
Counter Example- <br />
<br />
A= {1,2}<br />
<br />
B= {3,4,1}<br />
<br />
C= {1,3,5}<br />
<br />
D= {2,5,3}<br />
<br />
So, LHS evaluates to {1,2,3}<br />
<br />
Ans RHS to { 1,3,5}.<br />
<br />
So LHS # RHS. <br />
<br />
FEEL FREE TO CORRECT!Set Theory & Algebrahttp://gateoverflow.in/127522/pgee-2017?show=128409#a128409Sat, 06 May 2017 21:52:11 +0000Answered: MATRIX
http://gateoverflow.in/107919/matrix?show=128340#a128340
<p>Here $I3$ is the best example for such matrix.
<br>
<br>
$A = \begin{bmatrix} 1 & 0 & 0\\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}$
<br>
<br>
Here even number entries are 6 and $|A|=1$.We can't get more than 6 even element for this type of matrix.</p>
<p> </p>
<p>Hence,Maximum number of even entries in A is <strong>$6$.</strong></p>Set Theory & Algebrahttp://gateoverflow.in/107919/matrix?show=128340#a128340Sat, 06 May 2017 10:39:54 +0000Answered: is D36 distributive ?
http://gateoverflow.in/52980/is-d36-distributive?show=128281#a128281
You can simply check whether d36 is distributive or not by checking whether we can get it by product of distinct primes for example<br />
D30=2.3.5=30 possible d40=2.2.2.5 not possible as 2 came 3 times not distinct d36=2.3.2.3 not distributive as we did not get it through product of distinct primesSet Theory & Algebrahttp://gateoverflow.in/52980/is-d36-distributive?show=128281#a128281Fri, 05 May 2017 19:55:35 +0000Answered: GATE2005-7
http://gateoverflow.in/1349/gate2005-7?show=128270#a128270
<p>The matrix of transitive closure of a relation on a set of n elements</p>
<p>can be found using <strong>n<sup>2</sup>(2n-1)(n-1) + (n-1)n<sup>2</sup></strong> bit operations, which gives the time complexity of <strong>O(n<sup>4</sup>)</strong></p>
<p>But using <a rel="nofollow" href="https://www.google.com.sg/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwivopzL7tjTAhXHuI8KHfLMB7gQFgggMAA&url=http%3A%2F%2Fcs.winona.edu%2Flin%2Fcs440%2Fch08-2.pdf&usg=AFQjCNEZXhasfbdAghzoczS02ruseS4_KA">Warshall's Algorithm: Transitive Closure</a> we can do it in <strong>O(n<sup>3</sup>)</strong> bit operations</p>
<p>Hence <strong>D</strong> is the correct answer...</p>Set Theory & Algebrahttp://gateoverflow.in/1349/gate2005-7?show=128270#a128270Fri, 05 May 2017 19:06:20 +0000Answered: Discrete Mathematics Thegatebook
http://gateoverflow.in/128179/discrete-mathematics-thegatebook?show=128236#a128236
<p>Cartesian product of two set A and B is given as:</p>
<p><strong> AxB = { (a,b) | a$\epsilon$A $\Lambda$ b$\epsilon$B }</strong></p>
<p>Now we have A as <strong>empty set</strong> and B as <strong>non empty set</strong>. therefore there is no element<strong> a</strong> such that <strong>a$\epsilon$A </strong>satisfy ,due to which the condition <strong>a$\epsilon$A $\Lambda$ b$\epsilon$B</strong> fails....</p>
<p>so we can conclude that<strong> AxB</strong> would also be an empty set...</p>Set Theory & Algebrahttp://gateoverflow.in/128179/discrete-mathematics-thegatebook?show=128236#a128236Fri, 05 May 2017 15:23:33 +0000Answered: GATE2005-8
http://gateoverflow.in/1157/gate2005-8?show=128010#a128010
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=17444725974849341865"></p>
<p>we can also check by taking any three random sets</p>Set Theory & Algebrahttp://gateoverflow.in/1157/gate2005-8?show=128010#a128010Wed, 03 May 2017 22:07:36 +0000Answered: GATE2003-39
http://gateoverflow.in/930/gate2003-39?show=127960#a127960
<p><sup>Option B is correct</sup></p>
<p>n=3 (in options length is given as 3)</p>
<p>Let s1=a,s2=a,s3=a</p>
<p>now,</p>
<p>f(s1)=f(a)=2<sup>3 </sup></p>
<p>f(s2)=f(a)=2<sup>3 </sup></p>
<p>f(s3)=f(a)=2<sup>3 </sup></p>
<p>p1=2,p2=3,p3=5</p>
<p><strong>h(s1,s2,s3)=p1<sup>f(s1)</sup>p2<sup>f(s2)</sup>p3<sup>f(s3)</sup>=2<sup>8</sup>3<sup>8</sup>5<sup>8</sup></strong></p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/930/gate2003-39?show=127960#a127960Wed, 03 May 2017 15:49:03 +0000Sets and Algebra
http://gateoverflow.in/127198/sets-and-algebra
A = {5 , {6}, {7} }<br />
<br />
let P be the power set of A which of the following is true<br />
<br />
(1) ∅ ∊ P<br />
<br />
(2) ∅ ⊆ P<br />
<br />
(3) {5, {6}} ∊ P<br />
<br />
(4) {5, {6}} ⊆ PSet Theory & Algebrahttp://gateoverflow.in/127198/sets-and-algebraFri, 28 Apr 2017 09:51:58 +0000Answered: GATE2006-24
http://gateoverflow.in/987/gate2006-24?show=126771#a126771
<p>First let us understand what question is asking.</p>
<p>So π is a function from N to N, which just permutes the elements of N, so there will be n! such permutations.</p>
<p>Now given a particular π i.e. given a particular permutation scheme, we have to find number of permutations out of these n! permuations in which minimum elements of A and B after applying π to them are same.</p>
<p>So for example, if N = {1,2,3}, π is {2,3,1}, and if A is {1,3}, then π(A) = {2,1}. Now number of elements in A ∪ B is |A ∪ B|.</p>
<p>We can choose permutations for A ∪ B in nC|A∪B| ways. Note that here we are just choosing elements for permutation, and not actually permuting. Let this chosen set be P. Now once we have chosen numbers for permutations, we have to select mapping from each element of A ∪ B to some element of P. So first of all, to achieve required condition specified in question, we have to map minimum number in P to any of the number in A ∩ B, so that min(π(A)) = min(π(B)). We can do this in |A∩B| ways, since we can choose any element of |A∩B| to be mapped to minimum number in P. Now we come to permutation. We can permute numbers in P in |A∪B-1|! ways, since one number (minimum) is already fixed. Moreover, we can also permute remaining n - |A∪B-1| in (n - |A∪B-1|)! ways, so total no. of ways = nC|A∪B|∗|A∩B|∗|A∪B−1|!∗(n−|A∪B−1|)!=n!|A∩B||A∪B| So option (C) is correct. Note: Some answer keys on web have shown answer as option (D), which is clearly incorrect. Suppose |A ∪ B| = 3, and |A ∩ B| = 1, and n = 4, then option (D) evaluates to 14=0.25, which doesn't make sense. Source: <a rel="nofollow" href="http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html" target="_blank">http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html</a></p>Set Theory & Algebrahttp://gateoverflow.in/987/gate2006-24?show=126771#a126771Mon, 24 Apr 2017 09:25:51 +0000Answered: GATE2006-22
http://gateoverflow.in/983/gate2006-22?show=126617#a126617
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=14654609154771457889"></p>
<p><strong>Option C</strong> is ans</p>Set Theory & Algebrahttp://gateoverflow.in/983/gate2006-22?show=126617#a126617Sun, 23 Apr 2017 07:43:39 +0000Answered: GATE2003-31
http://gateoverflow.in/921/gate2003-31?show=126197#a126197
<p>‘a’ and ‘b’ are given as minimal elements. No other element in S is of lower order than either a or b.
<br>
‘c’ is given as maximum element. So, c is of higher order than any other element in S.
<br>
<br>
P(a) = True means all elements ‘x’ which have an edge from element ‘a’ have to be true.
<br>
Since there is an edge from ‘a’, we have to satisfy formula P(a) => P(x), which can only be done by setting
<br>
P(x) = True.
<br>
<br>
Elements which have an edge from b can be anything because formula P(b) => P(x) is satisfied as P(b) = False.
<br>
<br>
(A) This statement is true because making all elements true trivially satisfy formula P(x) => P(y).
<br>
<br>
(B) This statement is true if all elements are connected from b then all elements can be false.
<br>
<br>
(C) This statement is true because b<=x ensures x!=a and for all other elements P(x) can be false without violating the given implication.
<br>
(D) This statement is false. Since, P(a) = true , for all ‘x’ such that a<=x, P(x) must be true. We do have at least one such 'x', which is 'c' as it is the maximum element.
<br>
<br>
Thus, <strong>option (D) is the answer.</strong>
<br>
Source - <a rel="nofollow" href="http://quiz.geeksforgeeks.org/gate-gate-cs-2003-question-31/">http://quiz.geeksforgeeks.org/gate-gate-cs-2003-question-31/</a></p>Set Theory & Algebrahttp://gateoverflow.in/921/gate2003-31?show=126197#a126197Wed, 19 Apr 2017 14:21:02 +0000Answered: GATE2004-IT-4
http://gateoverflow.in/3645/gate2004-it-4?show=125560#a125560
<p>R1 is a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} . Under R1, an element x in A is related to an element y in B if x + y is divisible by 3.
<br>
Thus, R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}
<br>
R2 is a relation from B = {2, 4, 6, 8} to C = {1, 2, 3, 4} Under R2, an element y in B is related to an element z in C if y + z is even but not divisible by 3.
<br>
Thus, R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}
<br>
<a rel="nofollow" href="http://geeksquiz.com/wp-content/uploads/2016/01/it.jpg"><img alt="it" height="44" src="http://geeksquiz.com/wp-content/uploads/2016/01/it-300x44.jpg" width="300"></a>
<br>
Thus, R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
<br>
<br>
Thus, option (C) is correct. </p>Set Theory & Algebrahttp://gateoverflow.in/3645/gate2004-it-4?show=125560#a125560Fri, 14 Apr 2017 20:58:05 +0000Answered: GATE2004-24
http://gateoverflow.in/1021/gate2004-24?show=125546#a125546
Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R.<br />
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}. <br />
Now transitive closure is defined as smallest transitive relation which contains S. <br />
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now. <br />
<br />
Thus, option (B) matches the final set S.Set Theory & Algebrahttp://gateoverflow.in/1021/gate2004-24?show=125546#a125546Fri, 14 Apr 2017 17:49:10 +0000Answered: GATE2005-43
http://gateoverflow.in/1168/gate2005-43?show=125545#a125545
<p>A function f: X → Y is called on-to function if for every value in set Y, there is a value in set X.</p>
<p>Given that, f: B → C and g: A → B and h = f o g. </p>
<p>Note that the sign o represents <a rel="nofollow" href="https://en.wikipedia.org/wiki/Function_composition">composition</a>. </p>
<p>h is basically f(g(x)). So h is a function from set A
<br>
to set C.</p>
<p>It is also given that h is an onto function which means
<br>
for every value in C there is a value in A. </p>
<p>We map from C to A using B. So for every value in C, there must be a value in B. It means f must be onto. But g may or may not be onto as there may be some values in B which don't map to A. <strong>Example :</strong></p>
<p>Let us consider following sets
<br>
A : {a1, a2, a3}
<br>
B : {b1, b2}
<br>
C : {c1}</p>
<p>And following function values
<br>
f(b1) = c1
<br>
g(a1) = b1, g(a2) = b1, g(a3) = b1</p>
<p>Values of h() would be,
<br>
h(a1) = c1, h(a2) = c1, h(a3) = c1</p>
<p>Here h is onto, therefore f is onto, but g is
<br>
onto as b2 is not mapped to any value in A.</p>
<p>Given that, f: B → C and g: A → B and h = f o g.</p>Set Theory & Algebrahttp://gateoverflow.in/1168/gate2005-43?show=125545#a125545Fri, 14 Apr 2017 17:39:18 +0000Answered: Cyclic Group
http://gateoverflow.in/125469/cyclic-group?show=125474#a125474
<p>Let (G,*) be a <span class="marker">Cyclic group</span> of order ' n ': The number of <span class="marker">G</span>enerators is <span class="marker">G="Φ(n)"</span> </p>
<p><span class="marker">Euler's totient function</span> counts the positive integers up to a given integer n that are <a rel="nofollow" href="https://en.wikipedia.org/wiki/Relatively_prime">relatively prime</a> (co- prime) to n. </p>
<p> <span class="marker">Co-prime :</span> It can be defined more formally as the number of integers k in the range 1 ≤ <em>k</em> ≤ <em>n</em> for which the <a rel="nofollow" href="https://en.wikipedia.org/wiki/Greatest_common_divisor">greatest common divisor</a> gcd(<em>n</em>, <em>k</em>) is equal to 1. </p>
<p><span class="marker">For eg:</span> </p>
<p>Number of generators of cyclic group of order 3 = Φ(3) ={1,2} = 2 generators .</p>
<p>Number of generators of cyclic group of order 7 = Φ(7) = {1,2,3,4,5,6} = 6 generators .</p>
<p>Number of generators of cyclic group of order 6 = Φ(6) ={1,5} = 2 generators .</p>
<p>------------------------------------------------------------------------------------------------------------------</p>
<p><span class="marker">Suppose if the number is large then what will u do </span>: If n is very large then we need to do, split the n in such a way that it becomes multiplication of two prime numbers.</p>
<ul>
<li> n = p * q</li>
<li>Φ(n) = Φ(p) * Φ(q)</li>
</ul>
<p>for example: if we need to find out how many generators exists in cyclic group of order 77 then</p>
<p> 77 = 7 * 11</p>
<p> Φ(77) = Φ(7) * Φ(11)</p>
<p>By above explanation, Φ(7) = 6 generators and Φ(11) = 10 generators.</p>
<p>So total number of generators will be = 6 * 10 = 60 generators in cyclic group of order 77.</p>
<p>------------------------------------------------------------------------------------------------------------------</p>
<p>Eg 2: Number of generators in cyclic group of order 35:</p>
<p> Φ(35) = Φ(7) * Φ(5)</p>
<p> = 6 * 4 =24 generators.</p>
<p>------------------------------------------------------------------------------------------------------------------</p>
<p><span class="marker">Another special cases:</span> Number of generators in cyclic group of order 25:</p>
<p> Φ(25) = Φ(5<sup>2</sup>)</p>
<p><span class="marker">General Formula is:</span> if Φ(P<sup>n</sup>) = P<sup>n</sup> - P<sup>n-1</sup></p>
<p> Now Φ(25) = 5<sup>2</sup> - 5<sup>(2-1)</sup></p>
<p> = 20 generators.</p>
<p>------------------------------------------------------------------------------------------------------------------</p>
<p>Eg: Number of generators in cyclic group of order 84:</p>
<p> 84 = 2<sup>2</sup> * 3 * 7</p>
<p> Φ(84) = Φ(2<sup>2</sup> * 3 * 7)</p>
<p> = Φ(2<sup>2</sup>) * Φ(3) * Φ(7)</p>
<p> = 2<sup>2</sup> - 2<sup>(2-1)</sup> * 2 * 6</p>
<p> = 24 Generators</p>Set Theory & Algebrahttp://gateoverflow.in/125469/cyclic-group?show=125474#a125474Fri, 14 Apr 2017 09:38:56 +0000Answered: set theory
http://gateoverflow.in/122254/set-theory?show=125458#a125458
<p>Drawing hasse diagram for the poset ({{1}, {2}, {4},{1, 2}, {1, 4}, {2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}}, ⊆).</p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=4497923090195038245"></p>
<p>a) The maximal elements are the ones without any elements lying above them in the Hasse diagram, namely {1,2}, {1,3,4}, and {2,3,4}.</p>
<p>b) The minimal elements are the ones without any elements lying below them in the Hasse diagram, namely {1} , {2} ,and {4} .</p>
<p>c) There is no greatest element, since there is more than one maximal element, none of which is greater than the others.</p>
<p>d) There is no least element, since there is more than one minimal element, none of which is less than the others.</p>
<p>e) The upper bounds are the sets containing both {2} and {4} as subsets, i.e., the sets containing both 2 and 4 as elements. Pictorially, these are the elements lying above both {2} and {4} (in the sense of there being a path in the diagram), namely {2,4} and {2,3,4}.</p>
<p>f) The least upper bound is an upper bound that is less than every other upper bound. We found the upper bounds in part (e), and since {2,4} is less than (i.e., a subset of) {2,3,4}, we conclude that {2,4} is the least upper bound.</p>
<p>g) To be a lower bound of both {l, 3, 4} and {2, 3, 4}, a set must be a subset of each, and so must be a subset of their intersection, {3, 4}. There are only two such subsets in our poset, namely {3, 4} and { 4}. In the diagram, these are the points which lie below (in the path sense) both {1,3,4} and {2,3,4}.</p>
<p>h) The greatest lower bound is a lower bound that is greater than every other lower bound. We found the lower bounds in part (g), and since {3,4} is greater than (i.e., a superset of) {4}, we conclude that {3,4} is the greatest lower bound.</p>Set Theory & Algebrahttp://gateoverflow.in/122254/set-theory?show=125458#a125458Thu, 13 Apr 2017 21:05:58 +0000Answered: set theory
http://gateoverflow.in/122274/set-theory?show=125453#a125453
<p>In each case, we need to check whether every pair of elements has both a least upper bound and a greatest lower bound.</p>
<p>
<br>
<strong>a) This is a lattice.</strong> If we want to find the l.u.b. or g.l.b. of two elements in the same vertical column of the Hasse diagram, then we simply take the higher or lower (respectively) element.</p>
<p>If the elements are in different columns, then to find the g.l.b. we follow the diagonal line upward from the element on the left, and then continue upward on the right, if necessary to reach the element on the right.</p>
<p>For example, the l.u.b. of d and c is f; and the l.u.b. of a and e is e.</p>
<p>Finding greatest lower bounds in this poset is similar.</p>
<p><strong>b) This is not a lattice.</strong> Elements b and c have f, g, and h as upper bounds, but none of them is a l.u.b.</p>
<p><strong>c) This is a lattice.</strong> By considering all the pairs of elements, we can verify that every pair of them has a l.u.b. and a g.l.b.</p>
<p>For example, b and e have g and a filling these roles, respectively.</p>Set Theory & Algebrahttp://gateoverflow.in/122274/set-theory?show=125453#a125453Thu, 13 Apr 2017 20:45:09 +0000Answered: rosen(sets relation function)
http://gateoverflow.in/122683/rosen-sets-relation-function?show=125451#a125451
<p>If we write down the first few terms of this sum we notice a pattern.</p>
<p>For 1 to 3 GIF will be 1 ,</p>
<p>For 4 to 8 GIF will be 2 ,</p>
<p>For 9 to 15 GIF will be 3 ,</p>
<p>For 16 to 24 GIF will be 4</p>
<p>It starts (1 +1+1) + (2 + 2 + 2 +2 + 2) + (3 + 3 + 3 + 3 + 3 + 3 + 3) + · · ·. There are three l's, then five 2's, then seven 3's, and so on;
<br>
In general there are $(i+1)^{2} - i^{2}$ = 2i + 1 copies of i. So we need to sum i(2i + 1) for an appropriate range of values for i.</p>
<p>We must find this range. It gets a little messy at the end if m is such that the sequence stops before a complete range of the last value is present. Let n = floor(√m) - 1. Then there are n + 1 blocks, and $(n+1)^{2}$ - 1 is where the next-to-last block ends.</p>
<p>The sum of those complete blocks is
<br>
$\sum_{i=1}^{n}$ i(2i + 1) = $\sum_{i=1}^{n}$ 2$i^{2}$ + i = n(n + 1)(2n + 1)/3 + n(n + 1)/2.</p>
<p>The remaining terms in our summation all have the value n + 1 and the number of them present is m - ($(n + 1)^{2}$ - 1).</p>
<p><strong>Our final answer is therefore</strong>
<br>
n(n + 1)(2n + 1)/3 + n(n + 1)/2 + (n + 1)(m - $(n + 1)^{2}$ + 1).</p>Set Theory & Algebrahttp://gateoverflow.in/122683/rosen-sets-relation-function?show=125451#a125451Thu, 13 Apr 2017 20:28:05 +0000Answered: Rosen(relation)
http://gateoverflow.in/123123/rosen-relation?show=125446#a125446
a) The union of two relations is the union of these sets.<br />
<br />
Thus R1 U R2 holds between two integers if R1 holds or R2 holds (or both, it goes without saying).<br />
<br />
Thus (a, b) $\in$ R1 U R2 if and only if a ≡ b (mod 3) or a ≡ b (mod 4). There is not a good easier way to state this, other than perhaps to say that a - b is a multiple of either 3 or 4, or to work modulo 12 and write a - b ≡ 0, 3, 4, 6, 8, or 9 (mod 12).<br />
<br />
b) The intersection of two relations is the intersection of these sets.<br />
<br />
Thus R1 $\cap$ R2 holds between two integers if R1 holds and R2 holds.<br />
<br />
Thus (a, b) $\in$ R1 $\cap$ R2 if and only if a ≡ b (mod 3) and a ≡ b (mod 4). Since this means that a - b is a multiple of both 3 and 4, and that happens if and only if a - b is a multiple of 12, we can state this more simply as a ≡ b (mod 12).<br />
<br />
c) By definition R1 - R2 = R1 $\cap$ $\bar {R2}$ .<br />
<br />
Thus this relation holds between two integers if R1 holds and R2 does not hold.<br />
<br />
We can write this in symbols by saying that (a, b) $\in$ R1 - R2 if and only if a ≡ b (mod 3) and<br />
<br />
a $\not\equiv$ b (mod 4).<br />
<br />
d) By definition R2 - R1 = R2 $\cap$ $\bar {R1}$.<br />
<br />
Thus this relation holds between two integers if R2 holds and R1 does not hold.<br />
<br />
We can write this in symbols by saying that (a, b) $\in$ R2 - R1 if and only if a ≡ b (mod 4) and<br />
<br />
a $\not\equiv$ b (mod 3).<br />
<br />
e) We know that R1 $\oplus$ R2 = (R1 - R2) U (R2 -R1), so we look at our solutions to part (c) and part (d).<br />
<br />
Thus this relation holds between two integers if R1 holds and R2 does not hold, or vice versa.<br />
<br />
We can write this in symbols by saying that (a, b) $\in$ R1 $\oplus$ R2 if and only if (a ≡ b (mod 3) and<br />
<br />
a $\not\equiv$ b (mod 4)) or (a ≡ b (mod 4) and a $\not\equiv$ b (mod 3) ).<br />
<br />
We could also say that a - b is a multiple of 3 or 4 but not both.Set Theory & Algebrahttp://gateoverflow.in/123123/rosen-relation?show=125446#a125446Thu, 13 Apr 2017 19:25:02 +0000Answered: Rosen , Relations
http://gateoverflow.in/125415/rosen-relations?show=125436#a125436
<p>It is a theorem:</p>
<p><strong>Theorem :</strong>The relation R on a set A is transitive if and only if $R^{n}$ ⊆ R for n = 1, 2, 3, . . . .</p>
<p>Proof:</p>
<p>For if part suppose that $R^{n}$ ⊆ R for n = 1,2, 3, . . . .</p>
<p>In particular, $R^{2}$ ⊆ R.</p>
<p>To see that this implies R is transitive, note that if (a, b) ∈ R and (b, c) ∈ R, then by the definition of composition,</p>
<p>(a, c) ∈ $R^{2}$ .</p>
<p>Because $R^{2}$ ⊆ R, this means that (a, c) ∈ R. Hence, R is transitive.</p>
<p>Using mathematical induction to prove the only if part of the theorem.</p>
<p>Trivially true for n = 1.
<br>
Assume that $R^{n}$ ⊆ R, where n is a positive integer. </p>
<p>Assume that (a, b) ∈ $R^{n+1}$.</p>
<p>Then, because $R^{n+1}$ = $R^{n}$ ◦ R, there is an element y with y ∈ A such that (a, y) ∈ R and (y, b) ∈ $R^{n}$.</p>
<p>The inductive hypothesis, namely, that $R^{n}$ ⊆ R, implies that (y, b) ∈ R. Furthermore, because R is transitive, and (a, y) ∈ R
<br>
and (y, b) ∈ R, it follows that (a, b) ∈ R. This shows that $R^{n+1}$ ⊆ R.</p>
<p>Hence proved.</p>Set Theory & Algebrahttp://gateoverflow.in/125415/rosen-relations?show=125436#a125436Thu, 13 Apr 2017 18:20:17 +0000kenneith rosen
http://gateoverflow.in/125412/kenneith-rosen
At least how many numbers should be selected from the set {1, 5, 9, 13, …125} to be assured that two of the numbers selected have a sum of 146?Set Theory & Algebrahttp://gateoverflow.in/125412/kenneith-rosenThu, 13 Apr 2017 14:32:59 +0000#Big-O#Rosen
http://gateoverflow.in/124676/%23big-o%23rosen
Give Big-O estimate:<br />
<br />
${f(x)=n^{2n} + n^{n^2}}$<br />
<br />
The answer is given $O(n^{2n})$<br />
<br />
But, isn't $n^{n^2} > n^{2n}$ for n>2?<br />
<br />
If yes, then how is it $O(n^{2n})$?Set Theory & Algebrahttp://gateoverflow.in/124676/%23big-o%23rosenFri, 07 Apr 2017 18:43:29 +0000ISI 2004 MIII
http://gateoverflow.in/123963/isi-2004-miii
Q14 The inequality $\frac{2-gx+x^{2}}{1-x+x^{2}}\leq 3$ is true for all the value of x if and only if<br />
<br />
A) $1\leq g\leq 7$<br />
<br />
B) $-1\leq g\leq 1$<br />
<br />
C) $-6\leq g\leq 7$<br />
<br />
D) $-1\leq g\leq 7$Set Theory & Algebrahttp://gateoverflow.in/123963/isi-2004-miiiTue, 04 Apr 2017 15:03:00 +0000ISI 2004 MIII
http://gateoverflow.in/123882/isi-2004-miii
Q11 If $\alpha 1,\alpha 2,\dots,\alpha n$ are the positive numbers then<br />
<br />
$\frac{a1}{a2}+\frac{a2}{a3}+\dots+\frac{an-1}{an}+\frac{an}{a1}$<br />
<br />
is always<br />
<br />
A) $\geq n$<br />
<br />
B) $\leq n$<br />
<br />
C) $\leq n^{\frac{1}{2}}$<br />
<br />
D) None of the aboveSet Theory & Algebrahttp://gateoverflow.in/123882/isi-2004-miiiTue, 04 Apr 2017 10:11:48 +0000ISI 2004 MIII
http://gateoverflow.in/123819/isi-2004-miii
Q10 The equation $p\left ( x \right ) = \alpha$ where $p\left ( x \right ) = x^{4}+4x^{3}-2x^{2}-12x$ has four distinct real root if and only if<br />
<br />
A) $p\left ( -3 \right )<\alpha$<br />
<br />
B) $p\left ( -1 \right )>\alpha$<br />
<br />
C) $p\left ( -1 \right )<\alpha$<br />
<br />
D) $p\left ( -3 \right )<\alpha <p\left ( -1 \right )$Set Theory & Algebrahttp://gateoverflow.in/123819/isi-2004-miiiMon, 03 Apr 2017 22:44:41 +0000ISI 2004 MIII
http://gateoverflow.in/123818/isi-2004-miii
Q9 The equation'<br />
<br />
$\frac{1}{3}+\frac{1}{2}s^{2}+\frac{1}{6}s^{3}=s$<br />
<br />
has<br />
<br />
A) exactly three solution in [0.1]<br />
<br />
B) exactly one solution in [0,1]<br />
<br />
C) exactly two solution in [0,1]<br />
<br />
D) no solution in [0,1]Set Theory & Algebrahttp://gateoverflow.in/123818/isi-2004-miiiMon, 03 Apr 2017 22:40:09 +0000ISI 2004 MIII
http://gateoverflow.in/123813/isi-2004-miii
Q8 If $\alpha 1,\alpha 2,\alpha 3....\alpha n$ be the roots of $x^{n}+1=0$, then $\left ( 1-\alpha 1 \right )*\left ( 1-\alpha 2 \right )...\left ( 1-\alpha n \right )$ is equal to<br />
<br />
A) 1<br />
<br />
B) 0<br />
<br />
C) n<br />
<br />
D) 2Set Theory & Algebrahttp://gateoverflow.in/123813/isi-2004-miiiMon, 03 Apr 2017 21:54:38 +0000ISI 2004 MIII
http://gateoverflow.in/123770/isi-2004-miii
Q7 The equation $x^{6}-5x^{4}+16x^{2}-72x+9=0$ has<br />
<br />
A) Exactly two distinct real roots<br />
<br />
B) Exactly three distinct real roots<br />
<br />
C) Exactly four distinct real roots<br />
<br />
D) six different real rootsSet Theory & Algebrahttp://gateoverflow.in/123770/isi-2004-miiiMon, 03 Apr 2017 17:49:28 +0000ISI 2004 MIII
http://gateoverflow.in/123769/isi-2004-miii
Q6 If the equation $x^{4}+ax^{3}+bx^{2}+cx+1=0$ (where a,b,c are real number) has no real roots and if at least one of the root is of modulus one, then<br />
<br />
A) b=c<br />
<br />
B) a=c<br />
<br />
C) a=b<br />
<br />
D) none of thisSet Theory & Algebrahttp://gateoverflow.in/123769/isi-2004-miiiMon, 03 Apr 2017 17:45:54 +0000