GATE Overflow - Recent questions and answers in Set Theory & Algebra
http://gateoverflow.in/qa/mathematics/discrete-mathematics/set-theory-%26-algebra
Powered by Question2AnswerAnswered: $a_n = 4^n + 6^n$
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 +0000Answered: GATE 2016-1-2
http://gateoverflow.in/39636/gate-2016-1-2?show=127628#a127628
<p>The least value of 'n' for the recursion would be 3.</p>
<p>For n = 1, number of strings = 2 (0, 1)</p>
<p>For n = 2, number of strings = 3 (00, 01, 10)</p>
<p>For n = 3, number of strings = 5 (000, 001, 010, 100, 101)</p>
<p>For n = 4, number of strings = 8 (0000, 0001, 0010, 0100, 1000, 0101, 1010, 1001) ... </p>
<p>This seems to follow Fibonacci series and the recurrence relation for it is a<sub>n</sub> = a<sub>n−1</sub> + a<sub>n−2</sub>. Thus, B is the correct choice. </p>Set Theory & Algebrahttp://gateoverflow.in/39636/gate-2016-1-2?show=127628#a127628Mon, 01 May 2017 12:57:35 +0000Answered: GATE 2016-1-1
http://gateoverflow.in/39663/gate-2016-1-1?show=127626#a127626
(p ⇒ q) will give {8, 9, 10, 12} ¬r will give {8, 10, 11, 12} ¬s will give {8, 9, 10, 12} (¬r ∨ ¬s) will give {8, 9, 10, 11, 12} (p ⇒ q) ∧ (¬r ∨ ¬s) will give {8, 9, 10, 12} ¬((p ⇒ q) ∧ (¬r ∨ ¬s)) will give 11. Thus, C is the correct option.Set Theory & Algebrahttp://gateoverflow.in/39663/gate-2016-1-1?show=127626#a127626Mon, 01 May 2017 12:54:03 +0000Answered: GATE2015-3_5
http://gateoverflow.in/8399/gate2015-3_5?show=127383#a127383
{1, 1, 1, 1} {1, 1, 1, 2} {1, 1, 1, 3} {1, 1, 2, 2} {1, 1, 2, 3} {1, 1, 3, 3} {1, 2, 2, 2} {1, 2, 2, 3} {1, 2, 3, 3} {1, 3, 3, 3} {2, 2, 2, 2} {2, 2, 2, 3} {2, 2, 3, 3} {2, 3, 3, 3} {3, 3, 3, 3}<br />
<br />
so answer is 15Set Theory & Algebrahttp://gateoverflow.in/8399/gate2015-3_5?show=127383#a127383Sat, 29 Apr 2017 19:19:28 +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: 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 +0000Answered: kenneith rosen
http://gateoverflow.in/125412/kenneith-rosen?show=125432#a125432
<p>Another approach:A more general one though here since d sum is 146 which is a small value. Since the difference is 4 therefore all d numbers in d set can be obtained. Starting with 1+4=5,5+4=9,9+4=13,13+4=17,17+4=21,21+4=25,25+4=29,29+4=31,...and so on until d last value i.e. 125. Either by adding from front or subtracting from back all d no's. Like in my case I"ve summed up. When all the member no's of set r obtained then do d following:</p>
<p>The nos obtained r following:<strong>{1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81,85,89,93,97,101,105,109,113,117,121,125}</strong></p>
<p>From 146 start subtracting with d last no. Like 146-125=21, which is gain a member of d set. In a similar manner 146-121=25, both 21 and 25 r members of d set. Though it hardly took any time. Hence d following pairs were obtained:</p>
<p><strong>(125,21),(121,25),(113,33),(117,29),(109,37),(105,41),(101,45),(97,49),(93,53),(57,89),(85,61),(81,65),(77,69). </strong>These pairs total upto 146 which r 13 in no. If repetition not allowed, earlier mentioned excluding (73,73). </p>
<p>Final asn is 13. This approach is used for a smaller set. </p>Set Theory & Algebrahttp://gateoverflow.in/125412/kenneith-rosen?show=125432#a125432Thu, 13 Apr 2017 17:58:12 +0000Answered: Suppose a is a real number for which all the roots of the equation
http://gateoverflow.in/122625/suppose-is-real-number-for-which-all-the-roots-of-the-equation?show=125353#a125353
Option D should be Correct.<br />
<br />
1) Put x = 3/4 in given equation i.e $x^{4} - 2ax^{2} + x + a^{2} - a$ , the new equation will become<br />
<br />
$a^{2} - 2.125 a + 1.0664$<br />
<br />
then find the roots of this quadratic equation it will be.<br />
<br />
a= 0.812488 , a= 1.31251<br />
<br />
you can see both values of a are greater than 3/4.<br />
<br />
2) Now Put x = 2/3 in given equation i.e $x^{4} - 2ax^{2} + x + a^{2} - a$ the new equation will become<br />
<br />
$a^{2} - 1.8888 a + 0.864197$<br />
<br />
then find the roots of this quadratic equation.<br />
<br />
a= 0.777984 , a= 1.11082<br />
<br />
you can see both values of a are greater than 3/4<br />
<br />
Hence D option is correct.Set Theory & Algebrahttp://gateoverflow.in/122625/suppose-is-real-number-for-which-all-the-roots-of-the-equation?show=125353#a125353Thu, 13 Apr 2017 06:17:33 +0000Answered: Kenneth Rosen Edition7 Ch-1 Ex-1.2 QueNo-14
http://gateoverflow.in/42943/kenneth-rosen-edition7-ch-1-ex-1-2-queno-14?show=125181#a125181
<p>To look for hiking in West Virginia you could enter "<strong>HIKING AND (WEST AND VIRGINIA)</strong>"</p>
<p>Hiking in Virginia but not West Virginia could be entered by "<strong>(HIKING AND VIRGINIA) AND NOT WEST</strong>". Only saying "not west" is necessary and sufficient, to exclude <strong>West Virginia Hiking </strong>results from the larger set of <strong>Virginia hiking</strong> results. though larger search terms can give same results, but searching algorithms with minimum terms used are the norms.</p>Set Theory & Algebrahttp://gateoverflow.in/42943/kenneth-rosen-edition7-ch-1-ex-1-2-queno-14?show=125181#a125181Wed, 12 Apr 2017 06:30:40 +0000Answered: Kenneth Rosen Edition7 Ch-1 Ex-1.2 QueNo-13
http://gateoverflow.in/42929/kenneth-rosen-edition7-ch-1-ex-1-2-queno-13?show=125180#a125180
To search for beaches in New Jersey you could type NEW AND JERSEY AND BEACHES. If you enter (JERSEY AND BEACHES) NOT NEW, then you'll get sites about beaches on the isle of Jersey. If it was known that the word "isle" was in the name of the location, then you would enter ISLE AND JERSEY AND BEACHES.Set Theory & Algebrahttp://gateoverflow.in/42929/kenneth-rosen-edition7-ch-1-ex-1-2-queno-13?show=125180#a125180Wed, 12 Apr 2017 06:18:28 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123769/isi-2004-miii?show=124989#a124989
<p>Option D Should be Correct</p>
<p> Alternate forms of equation:</p>
<p> </p>
<p><img alt="x^3 (a + x) + b x^2 + c x + 1" height="18" src="http://www5b.wolframalpha.com/Calculate/MSP/MSP62520h66165i4b4306g000041i6i2021f325cah?MSPStoreType=image/gif&s=61" width="159"></p>
<p> </p>
<p><img alt="a x^3 + x^2 (b + x^2) + c x + 1" height="22" src="http://www5b.wolframalpha.com/Calculate/MSP/MSP62720h66165i4b4306g00000eig2i7b621bf571?MSPStoreType=image/gif&s=61" width="165"></p>
<p>Above equation has no real roots and if at least one of the root is of modulus one (which is both +1 and -1) as stated by question then it should be integer root of the equation.
<br>
</p>
<p>Putting x= 1 in the main equation we get : c = - a - b - 2............(1)</p>
<p>Putting x= -1 in the main equation we get : c = - a + b + 2.............(2)</p>
<p>Adding (1) and (2) we get</p>
<p>2c = - 2a</p>
<p>c= -a</p>
<p>Correct me if wrong</p>
<p> </p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/123769/isi-2004-miii?show=124989#a124989Mon, 10 Apr 2017 10:57:57 +0000Answered: #Big-O#Rosen
http://gateoverflow.in/124676/%23big-o%23rosen?show=124821#a124821
Yes it should be O (n^n^2)Set Theory & Algebrahttp://gateoverflow.in/124676/%23big-o%23rosen?show=124821#a124821Sat, 08 Apr 2017 23:56:54 +0000Answered: ISRO2014-50
http://gateoverflow.in/54987/isro2014-50?show=124318#a124318
Answer is ASet Theory & Algebrahttp://gateoverflow.in/54987/isro2014-50?show=124318#a124318Wed, 05 Apr 2017 21:26:06 +0000Answered: composition of function
http://gateoverflow.in/108396/composition-of-function?show=124271#a124271
all are validSet Theory & Algebrahttp://gateoverflow.in/108396/composition-of-function?show=124271#a124271Wed, 05 Apr 2017 18:29:27 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123819/isi-2004-miii?show=124235#a124235
<h2>Alternate form of equation is:</h2>
<p><img alt="x (x + 2) (x^2 + 2 x - 6) = 0" src="http://www4f.wolframalpha.com/Calculate/MSP/MSP3971dc41gih41b1hd58000050h250g2fdcfi4ha?MSPStoreType=image/gif&s=26"></p>
<p>or </p>
<p><img alt="x (x^3 + 4 x^2 - 2 x - 12) = 0" src="http://www4f.wolframalpha.com/Calculate/MSP/MSP3991dc41gih41b1hd58000048a92dc4idigfd3b?MSPStoreType=image/gif&s=26"></p>
<p>As p(x) = α</p>
<p>Roots of the equation or value of α are</p>
<p>x = - 2</p>
<p>x = 0</p>
<p>$x = - 1 - \sqrt{7}$</p>
<p>$x = \sqrt{7} -1$</p>
<p>p(-3) = - 9</p>
<p>p(-1) = 7</p>
<p>Hence option D is correct.</p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/123819/isi-2004-miii?show=124235#a124235Wed, 05 Apr 2017 15:39:43 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123770/isi-2004-miii?show=124106#a124106
<p>P(x) = x<sup>6 </sup>− 5x<sup>4 </sup>+ 16x <sup>2 </sup>− 72x + 9 </p>
<ul>
<li>P(0) = 9</li>
<li>P(-1)= -ve</li>
<li>P(4) = +ve</li>
</ul>
<p>Hence, atleast 2 real roots can be clearly seen, but what about other 4 roots left ?</p>
<p>That's why we check P''(x),</p>
<p>P'(x) = 6x<sup>5 </sup>- 20x<sup>3 </sup>+ 32x - 72</p>
<p>P''(x) = 30x <sup>4</sup>- 60x<sup>2 </sup>+ 32 > 0 for any real value of x.</p>
<p>comparing this with a quadratic eq taking x<sup>2</sup> as y we get 30y<sup>2 </sup>- 60y + 32. The discriminant (b<sup>2 </sup>- 4ac) is negative implying P''(x) has no real roots</p>
<p>Hence by Rolle's theorem P'(x) can have at most 1 real root and P(x) can have at most 2 real roots. Because if a function f(x) has 2 roots x<sub>1</sub> and x<sub>2</sub> then there exists a point x ∈ [x<sub>1</sub>, x<sub>2</sub>] where the curve becomes flat i.e its the root of f'(x) meaning f(x) has max 2 roots</p>
<p> => <strong>Exactly two distinct real roots</strong></p>
<p><strong>Option A</strong></p>Set Theory & Algebrahttp://gateoverflow.in/123770/isi-2004-miii?show=124106#a124106Wed, 05 Apr 2017 00:46:56 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123730/isi-2004-miii?show=124090#a124090
Keep n=3 it would come out to be 3*(x^(2)) +3*(x) so it would be factor of x^(2) +(x)+1Set Theory & Algebrahttp://gateoverflow.in/123730/isi-2004-miii?show=124090#a124090Tue, 04 Apr 2017 23:17:47 +0000Answered: GATE2017-2-21
http://gateoverflow.in/118278/gate2017-2-21?show=124058#a124058
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=17031793655642334638"></p>
<p>hope it might help.......</p>Set Theory & Algebrahttp://gateoverflow.in/118278/gate2017-2-21?show=124058#a124058Tue, 04 Apr 2017 20:52:27 +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
<p>Q11 If $\alpha 1,\alpha 2,...\alpha n$ are the positive numbers then</p>
<p>$\frac{a1}{a2}+\frac{a2}{a3}....\frac{an-1}{an}+\frac{an}{a1}$</p>
<p>is always</p>
<p>A) $\geq n$</p>
<p>B) $\leq n$</p>
<p>C) $\leq$n<sup>$\frac{1}{2}$</sup></p>
<p>D) None of the above</p>Set Theory & Algebrahttp://gateoverflow.in/123882/isi-2004-miiiTue, 04 Apr 2017 10:11:48 +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 +0000Rosen(relations)
http://gateoverflow.in/123320/rosen-relations
Given the directed graphs representing two relations, how can the directed graph of the union, intersection, symmetric difference, difference, and composition of these relations be found?Set Theory & Algebrahttp://gateoverflow.in/123320/rosen-relationsSat, 01 Apr 2017 19:38:35 +0000Rosen(relations)
http://gateoverflow.in/123319/rosen-relations
Let R be a relation on a set A. Explain how to use the directed<br />
graph representing R to obtain the directed graph<br />
representing the complementary relation $\overline{R}$Set Theory & Algebrahttp://gateoverflow.in/123319/rosen-relationsSat, 01 Apr 2017 19:37:24 +0000ISI 2017
http://gateoverflow.in/123274/isi-2017
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=13525049551868792309"></p>
<p>I find alpha <x+y which gives me x+y<2. But the answer is A. Can someone please help.</p>Set Theory & Algebrahttp://gateoverflow.in/123274/isi-2017Sat, 01 Apr 2017 13:45:57 +0000Rosen (relation)
http://gateoverflow.in/123125/rosen-relation
How many relations are there on the set {a, b, c, d}<br />
that contain the pair (a, a)?Set Theory & Algebrahttp://gateoverflow.in/123125/rosen-relationFri, 31 Mar 2017 17:27:49 +0000Suppose a, b, c > 0 are in geometric progression and a
http://gateoverflow.in/122639/suppose-a-b-c-0-are-in-geometric-progression-and-a
<p>Suppose a; b; c > 0 are in geometric progression and a<sup>p</sup> = b<sup>q</sup> = c<sup>r</sup> != 1.
<br>
Which one of the following is always true?
<br>
(A) p, q, r are in geometric progression
<br>
(B) p, q, r are in arithmetic progression
<br>
(C) p, q, r are in harmonic progression
<br>
(D) p = q = r</p>Set Theory & Algebrahttp://gateoverflow.in/122639/suppose-a-b-c-0-are-in-geometric-progression-and-aSun, 26 Mar 2017 23:22:44 +0000