GATE Overflow - Recent questions and answers in Set Theory & Algebra
http://gateoverflow.in/qa/mathematics/discrete-mathematics/set-theory-%26-algebra
Powered by Question2AnswerSuppose 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
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=13394253793924842238"></p>Set Theory & Algebrahttp://gateoverflow.in/122625/suppose-is-real-number-for-which-all-the-roots-of-the-equationSun, 26 Mar 2017 16:01:43 +0000Answered: GATE 2016-1-27
http://gateoverflow.in/39714/gate-2016-1-27?show=122589#a122589
<p>watch step by step soln. of above problem @ <a rel="nofollow" href="https://www.youtube.com/watch?v=BWOVko_y1xQ&t=25s">https://www.youtube.com/watch?v=BWOVko_y1xQ&t=25s</a></p>Set Theory & Algebrahttp://gateoverflow.in/39714/gate-2016-1-27?show=122589#a122589Sun, 26 Mar 2017 07:10:36 +0000Answered: set theory
http://gateoverflow.in/122391/set-theory?show=122576#a122576
<p>There are two necessary and sufficient conditions for a POSET to be boolean algebra :</p>
<p>1. Number of elements should be 2^n.</p>
<p>2.Number of edges should be n*2^(n-1).</p>
<p>In given diagram,these both conditions are true and only those who are isomorphic to this graph will be boolean algebra.</p>
<p><strong>Option A is correct</strong></p>Set Theory & Algebrahttp://gateoverflow.in/122391/set-theory?show=122576#a122576Sun, 26 Mar 2017 05:22:08 +0000Answered: group theory
http://gateoverflow.in/122528/group-theory?show=122575#a122575
Since a subgroup is also a group,it must satisfy all the properties of group..not only algebraic structure.Set Theory & Algebrahttp://gateoverflow.in/122528/group-theory?show=122575#a122575Sun, 26 Mar 2017 05:15:02 +0000Answered: group theory
http://gateoverflow.in/122536/group-theory?show=122545#a122545
order of subgroup divides order of group.Set Theory & Algebrahttp://gateoverflow.in/122536/group-theory?show=122545#a122545Sat, 25 Mar 2017 13:48:02 +0000Answered: group theory
http://gateoverflow.in/122494/group-theory?show=122502#a122502
<p>The answer will be A. Only $S1$ is the group.
<br>
<br>
Because in the second case, Identity element does not exist.
<br>
<br>
In the first case identity element is $0$. That means for all $ a \in S1 $, $ a +_{m} 0 = a $. because $a< m$
<br>
<br>
In the second case, $0$ can not be the identity element. For example: for one of the member $m$ of the set we have $ m +_{m} 0 = 0 $ , It should come $m$. That's why $S2$ is not a group.</p>
<p>However, both $S1$ and $S2$ are <span style="background-color:#ffff00">Semigroup </span>as they satisfy <span class="marker">closure</span> and <span class="marker">associativity</span> property.</p>Set Theory & Algebrahttp://gateoverflow.in/122494/group-theory?show=122502#a122502Sat, 25 Mar 2017 05:18:15 +0000Answered: group theory
http://gateoverflow.in/122470/group-theory?show=122490#a122490
<blockquote>
<p><strong>Group:</strong> For any algebraic structure to be a group, that has to satisfy the Closure, Associatively, Identity and Inverse properties. </p>
</blockquote>
<p><strong>Closure</strong></p>
<p><strong>For all $a, b \in G$, the result of the operation, $a * b$, is also in $G$.</strong></p>
<p>In above example, Since there is one element, hence $ a = b = 0 $, and $ a * b = 0 * 0 = 0 \in G $</p>
<p>Hence Closure satisfy.</p>
<p><strong>Associative</strong></p>
<p>For all $a, b, c \in G$, $ (a * b) * c = a * (b * c) $.</p>
<p>For above example, $ a = b = c = 0 $</p>
<p>Hence $ (a * b) * c = a * (b * c) $</p>
<p> $ \implies (0 * 0) * 0 = 0 * (0 * 0) \implies 0 = 0 $</p>
<p>Hence Associatively satisfied.</p>
<p><strong>Identity element</strong></p>
<p>There exists an element $e \in G$ such that, for every element $a \in G$, the equation $ e * a = a * e = a $ holds. Such an element is unique, and thus one speaks of the identity element.</p>
<p>For above example $ a = e = 0 $ </p>
<p>Hence $ e * a = a * e \implies 0 * 0 = 0 * 0 \implies 0 = a $</p>
<p>Hence $ e = 0 $ is the identity element. </p>
<p><strong>Inverse element</strong></p>
<p>For each $a \in G$, there exists an element $b \in G$, commonly denoted $a^{−1}$, such that $a * b = b * a = e $, where $e$ is the identity element.</p>
<p>For your example, The inverse element is $0$. Because when you multiply $0$ with $0$ then you will get $0$, which is also an identity element of the structure. </p>Set Theory & Algebrahttp://gateoverflow.in/122470/group-theory?show=122490#a122490Sat, 25 Mar 2017 03:30:03 +0000Answered: set theory
http://gateoverflow.in/122378/set-theory?show=122449#a122449
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=9310623582846890055"></p>
<p>solution</p>Set Theory & Algebrahttp://gateoverflow.in/122378/set-theory?show=122449#a122449Fri, 24 Mar 2017 08:41:39 +0000Answered: set theory
http://gateoverflow.in/122284/set-theory?show=122413#a122413
You're correct about A,B,C<br />
<br />
A) Not Lattice<br />
<br />
B & C) Totally ordered set<br />
<br />
D) A Lattice with LUB(a,b) = a UNION b & GLB(a,b) = a INTERSECTION b.<br />
<br />
for e.g LUB{(1,2), (2,3)} = {1,2,3} & GLB{(1,2), (2,3)} = {2}Set Theory & Algebrahttp://gateoverflow.in/122284/set-theory?show=122413#a122413Thu, 23 Mar 2017 18:26:16 +0000Answered: set theory
http://gateoverflow.in/122322/set-theory?show=122333#a122333
Definitely true, if lattice is finite then it has definitely some fix upper bound and lower bound, If we have given finiteness in lattice then it must be bounded...Set Theory & Algebrahttp://gateoverflow.in/122322/set-theory?show=122333#a122333Wed, 22 Mar 2017 14:10:34 +0000Answered: set theory
http://gateoverflow.in/122325/set-theory?show=122326#a122326
<p>please verify
<br>
<br>
B) and C) are not partial orders itself,since they are not reflexive.So they cant be lattices and so cant be distributive lattices.
<br>
<br>
A) is a distributive lattices,because </p>
<p> In A) LUB is nothing but UNION operation and GLB is nothing but INTERSECTION operation and we know that "<em><strong>union is distributive over intersection and intersection is distributive over union" </strong></em>from set theory. So we can say that LUB is distributive over GLB and vice-versa.</p>
<p>B) is a distributive lattice because</p>
<p> In B) LUB is nothing but INTERSECTION operation and GLB is nothing but UNION operation.Again from set theory we already know that "<em><strong>union is distributive over intersection and intersection is distributive over union". </strong></em>
<br>
<br>
Also if someone can provide some more proofs for A) and D) to be called as distributive lattices,it will be helpful.
<br>
<br>
<br>
<br>
.</p>Set Theory & Algebrahttp://gateoverflow.in/122325/set-theory?show=122326#a122326Wed, 22 Mar 2017 12:00:41 +0000set theory
http://gateoverflow.in/122314/set-theory
<p><em><strong>"Every sub-lattice of a distributive lattice is distributive"</strong></em>. can somebody prove it ??</p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/122314/set-theoryWed, 22 Mar 2017 07:17:40 +0000set theory
http://gateoverflow.in/122289/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=17465271246629359046"></p>Set Theory & Algebrahttp://gateoverflow.in/122289/set-theoryTue, 21 Mar 2017 12:45:28 +0000set theory
http://gateoverflow.in/122274/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=18191037133881223458"></p>
<p>A) lattice </p>
<p>B) not a lattice since b and c have f,g,h as upper bounds but none of the three is least (i.e) one is not lesser than other 2. so b and c dont have a least upper bound.</p>
<p>C) lattice </p>Set Theory & Algebrahttp://gateoverflow.in/122274/set-theoryTue, 21 Mar 2017 09:20:59 +0000set theory
http://gateoverflow.in/122271/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=12234657340731390463"></p>Set Theory & Algebrahttp://gateoverflow.in/122271/set-theoryTue, 21 Mar 2017 09:09:22 +0000Answered: set theory
http://gateoverflow.in/122250/set-theory?show=122258#a122258
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=4837128389851446701"></p>
<p>verify pls..</p>Set Theory & Algebrahttp://gateoverflow.in/122250/set-theory?show=122258#a122258Tue, 21 Mar 2017 07:57:43 +0000set theory
http://gateoverflow.in/122254/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=15205686062765178900"></p>Set Theory & Algebrahttp://gateoverflow.in/122254/set-theoryTue, 21 Mar 2017 07:15:11 +0000Answered: set theory
http://gateoverflow.in/122247/set-theory?show=122251#a122251
<p>a) <em>l</em> and <em>m</em> are the maximal elements as they have no successor</p>
<p>b) <em>a</em>, <em>b</em> and <em>c </em>are the minimal elements as they have no predecessor</p>
<p>c) There is no greatest elements as the 2 maximal elements are incomparable</p>
<p>d) Same as above, <em>a</em>, <em>b</em> and <em>c</em> are minimal and incomparable</p>
<p>e) Upper bounds : <em>a</em>={<em>a,d,i,j,l,h,k,m</em>} <em>b</em>={<em>b,e,h,k,m,d,i,j,l</em>} c={<em>c,f,g,k,m,l</em>} (<em>a,b,c</em>)={<em>k,m,l</em>}</p>
<p>f) Least upper bound of {a,b,c} is <em>k</em></p>
<p>g) Lower bounds : <em>f</em>={<em>f,c</em>} <em>g</em>={<em>g,f,c</em>} <em>h</em>={<em>h,e,b,d,a</em>}</p>
<p>h) Greatest lower bound of {<em>f,g,h</em>} does not exist as they have no lower bounds in common</p>Set Theory & Algebrahttp://gateoverflow.in/122247/set-theory?show=122251#a122251Tue, 21 Mar 2017 06:46:39 +0000Answered: set theory
http://gateoverflow.in/122223/set-theory?show=122231#a122231
YES, Since friendship is symmetric relation i.e. A->B (A is friend of B) then B->A also (B is friend of A) and lets say A and B both have common friend C then we have A->C, C->A, B->C and C->B, from this { A->B, B->A, A->C, C->A, B->C, C->B} we can infer that it is not transitive relation.Set Theory & Algebrahttp://gateoverflow.in/122223/set-theory?show=122231#a122231Mon, 20 Mar 2017 18:33:38 +0000Answered: set theory
http://gateoverflow.in/122224/set-theory?show=122230#a122230
I think answer option (A) is correct because we can eliminate option D because it is not following transitive relation, option C does not follow anti symmetry relation and option A does not follow reflexive relation.Set Theory & Algebrahttp://gateoverflow.in/122224/set-theory?show=122230#a122230Mon, 20 Mar 2017 18:24:39 +0000set theory
http://gateoverflow.in/122212/set-theory
<p>what is the difference between <em><strong>maximum/minimum</strong></em> AND <em><strong>greatest/least </strong></em>AND <em><strong>upperbound/lowerbound</strong></em></p>Set Theory & Algebrahttp://gateoverflow.in/122212/set-theoryMon, 20 Mar 2017 13:14:27 +0000Answered: set theory
http://gateoverflow.in/121897/set-theory?show=122204#a122204
A and D are equivalence relations<br />
<br />
B is not transitive<br />
<br />
C is not reflexive, not symmetric and not transitive<br />
<br />
E is neither reflexive nor transitiveSet Theory & Algebrahttp://gateoverflow.in/121897/set-theory?show=122204#a122204Mon, 20 Mar 2017 12:03:33 +0000groups in DM
http://gateoverflow.in/122083/groups-in-dm
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=16397532371912851652"></p>Set Theory & Algebrahttp://gateoverflow.in/122083/groups-in-dmSun, 19 Mar 2017 11:31:07 +0000Answered: IIITH-PGEE
http://gateoverflow.in/121819/iiith-pgee?show=122021#a122021
<p>6^4 because for every candy we have 6 choices (Permutation, Here order matters )</p>
<p>(6+4-1) C<sub> 4</sub> = 126 (Combination with repetition, Here order doesn't matter )</p>Set Theory & Algebrahttp://gateoverflow.in/121819/iiith-pgee?show=122021#a122021Sat, 18 Mar 2017 15:42:40 +0000Answered: set theory
http://gateoverflow.in/121775/set-theory?show=121922#a121922
<p><strong>a) and b)</strong></p>
<p><strong><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=17311712068742674488"></strong></p>
<p> </p>
<p>No of relations in both (a) and (b) = $\begin{align*} 2^{\left ( n^2 - 1 \right )} \end{align*}$</p>
<p> </p>
<p>(c) </p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=13156901957137182677"></p>
<p> </p>
<p>No of relations such that no pair in $R$ has $a$ as its first element = $\begin{align*} 2^{\left ( n^2 - n \right )} \end{align*}$</p>
<p> </p>
<p>(d)</p>
<p>No of relations such that at least one ordered pair in $R$ has $a$ as its first element = $\begin{align*} 2^{\left ( n^2 \right )} - 2^{\left ( n^2 - n \right )} \end{align*}$</p>
<p> </p>
<p><strong>(e)</strong></p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=15082800710892741258"></p>
<p> </p>
<p>Here we are not allowing a as first element or b as second element of any pair ($x$,$y$) $\in$ $R$</p>
<p>No of such relation = $\begin{align*} 2^{n^2 - \left ( 2n-1 \right )} \end{align*}$</p>
<p><strong>(f)</strong></p>
<p>Here we must have atleast one pair ($x$,$y$) such that $a$ is the first element or $b$ is second element.</p>
<p>No of such relation = $\begin{align*} 2^{n^2} - 2^{n^2 - \left ( 2n-1 \right )} \end{align*}$</p>Set Theory & Algebrahttp://gateoverflow.in/121775/set-theory?show=121922#a121922Fri, 17 Mar 2017 11:27:26 +0000set theory
http://gateoverflow.in/121901/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=10960471889966698538"></p>Set Theory & Algebrahttp://gateoverflow.in/121901/set-theoryFri, 17 Mar 2017 08:28:18 +0000set theory
http://gateoverflow.in/121809/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=6507082804701728388"></p>Set Theory & Algebrahttp://gateoverflow.in/121809/set-theoryThu, 16 Mar 2017 10:12:57 +0000set theory
http://gateoverflow.in/121803/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=8423156958859164858"></p>Set Theory & Algebrahttp://gateoverflow.in/121803/set-theoryThu, 16 Mar 2017 08:51:56 +0000Answered: set theory
http://gateoverflow.in/121752/set-theory?show=121753#a121753
To find So R we want to find the set of pairs (a, c) such that for some person b, a is a parent of b, and b<br />
is a sibling of c. Since brothers and sisters have the same parents, this means that a is also the parent of c.<br />
Thus S o R is contained in the relation R. More specifically, (a, c) E S o R if and only if a is the parent of c,<br />
and c has a sibling (who is necessarily also a child of a). To find Ro S we want to find the set of pairs (a, c)<br />
such that for some person b, a is a sibling of b, and b is a parent of c. This is the same as the condition that<br />
a is the aunt or uncle of c .Set Theory & Algebrahttp://gateoverflow.in/121752/set-theory?show=121753#a121753Thu, 16 Mar 2017 06:35:06 +0000Answered: set theory
http://gateoverflow.in/121715/set-theory?show=121716#a121716
<p>26) A) R<strong><sup>-1</sup></strong> = { (b,a) iff (a<b) }
<br>
<br>
B) R<strong><sup>c</sup></strong> = { (a,b) iff (a>=b) }
<br>
<br>
27) A) R<sup><strong>-1</strong></sup> = { (b,a) iff a divides b }
<br>
<br>
B) R<strong><sup>c</sup></strong> = { (a,b) iff a doesnot divide b }</p>Set Theory & Algebrahttp://gateoverflow.in/121715/set-theory?show=121716#a121716Wed, 15 Mar 2017 16:23:35 +0000Answered: set theory
http://gateoverflow.in/121676/set-theory?show=121679#a121679
<p><strong>Reflexive</strong> as x will always be a multiple of x.</p>
<p><strong>Not symmetric</strong>. because (4,2) ∈ R but (2,4) doesnot ∈ R</p>
<p><strong>Not Anti-symmetric</strong> as (-5,5) ∈ R and (5,-5) ∈ R. <em><strong>Note :</strong></em> (-5) is a multiple of 5 as <strong><em>5*(-1) = (-5)</em></strong> and 5 is a multiple of (-5) as<em><strong> (-5)*(-1) = 5</strong></em>.</p>
<p><strong>Transitive</strong> Because if (x,y) ∈ R, then x = y*p and if (y,z) ∈ R,then y = z*q. Now (x,z) will also be in R as x is a multiple of z.Because x = y*p = (z*q)*p = (p*q)*z.</p>Set Theory & Algebrahttp://gateoverflow.in/121676/set-theory?show=121679#a121679Wed, 15 Mar 2017 12:30:18 +0000Answered: set theory
http://gateoverflow.in/121674/set-theory?show=121675#a121675
<p>we say that x $\equiv$ y(mod n) iff n divides (x-y) <em><strong>(i.e) </strong></em>(x-y) is a multiple of 7.
<br>
<br>
here relation is x $\equiv$ y(mod 7) which means (x-y) = 7m where m is some integer.
<br>
<br>
If (x-y) = 7m then (y-x) = -7m.so i can also write as y $\equiv$ x(mod 7).
<br>
<br>
So if (x,y) $\in$ R , then (y,x) $\in$ R.<strong>So this relation is symmetric. Also this shows this relation is not anti-symmetric.</strong>
<br>
<br>
<strong>This relation is reflexive</strong> because x $\equiv$ x(mod n) because (x-x) = 0 which is divisible by 7.
<br>
<br>
<strong>This relation is transitive</strong> because
<br>
<br>
if (x,y) $\in$ R, then x $\equiv$ y(mod 7) <strong><em>(i.e)</em></strong> (x-y) = 7m ---->1
<br>
<br>
if (y,z) $\in$ R, then y $\equiv$ z(mod 7) <em><strong>(i.e)</strong></em> (y-z) = 7p.---->2
<br>
<br>
1) + 2) will give (x-z) = 7m+7p =7(m+p).so since (x-z) is also a multiple of 7, x $\equiv$ z(mod 7), which means (x,z) $\in$ R.</p>Set Theory & Algebrahttp://gateoverflow.in/121674/set-theory?show=121675#a121675Wed, 15 Mar 2017 11:44:33 +0000set theory
http://gateoverflow.in/121667/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=3364922567566726988"></p>Set Theory & Algebrahttp://gateoverflow.in/121667/set-theoryWed, 15 Mar 2017 10:30:30 +0000set theory
http://gateoverflow.in/121648/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=18152868624141192058"></p>Set Theory & Algebrahttp://gateoverflow.in/121648/set-theoryWed, 15 Mar 2017 07:26:51 +0000set theory
http://gateoverflow.in/121646/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=6036205878999374637"></p>Set Theory & Algebrahttp://gateoverflow.in/121646/set-theoryWed, 15 Mar 2017 07:21:00 +0000Answered: set theory
http://gateoverflow.in/121637/set-theory?show=121643#a121643
A) S<br />
<br />
B) R,S,T<br />
<br />
C) R,S <br />
<br />
D) A<br />
<br />
E) R,S,T<br />
<br />
F) S. NOT TRANSITIVE because 1*0 = 0 ,0*1 = 0 but 1*1 != 0<br />
<br />
G) A,T<br />
<br />
H) S NOT TRANSITIVE because (2,1) and (1,2) belongs to R but (2,2) doesnot belong to RSet Theory & Algebrahttp://gateoverflow.in/121637/set-theory?show=121643#a121643Wed, 15 Mar 2017 06:54:56 +0000Answered: set theory
http://gateoverflow.in/121632/set-theory?show=121635#a121635
<p><span class="marker">A)</span> <strong>R = { (0,1),(1,0),(1,1),(1,2),(1,3),(2,1),(2,3),(3,1),(3,2),(4,1),(4,3) }.</strong>
<br>
gcd (0,1) = 1 because 1 divides both 1 and 0 whereas gcd(0,2) = 2 because 2 divides both 0 and 2.
<br>
<span class="marker">B)</span> <strong>R = { (1,2),(2,1),(2,2) }.</strong>
<br>
There is no LCM of 0 and any non-zero integers (i.e) undefined.Because LCM(a,b) is "the smallest positive number m for which there exist positive integers (na)(a) and (nb)(b) such that (na)(a)=(nb)(b)=m".</p>Set Theory & Algebrahttp://gateoverflow.in/121632/set-theory?show=121635#a121635Wed, 15 Mar 2017 05:53:16 +0000Answered: set theory
http://gateoverflow.in/121622/set-theory?show=121623#a121623
if a divides b , then b = (a)(m) ---->1<br />
and if b divides c , then c = (b)(r) ---->2 <br />
from 1 and 2 , c = (b)(r) <br />
= (a)(m)(r) , which means a divides cSet Theory & Algebrahttp://gateoverflow.in/121622/set-theory?show=121623#a121623Wed, 15 Mar 2017 03:18:35 +0000Answered: set theory
http://gateoverflow.in/121620/set-theory?show=121621#a121621
R1,R2,R3,R4 are transuitive relationsSet Theory & Algebrahttp://gateoverflow.in/121620/set-theory?show=121621#a121621Wed, 15 Mar 2017 03:06:57 +0000Answered: set theory and algebra
http://gateoverflow.in/121592/set-theory-and-algebra?show=121612#a121612
<blockquote>
<p>for these type of question try to create 1 counter example if u succeed then then given statement is not correct else correct</p>
</blockquote>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=4428592448223986692"> </p>
<p>for these two question it is not possible to disprove using counter example implies that both are correct</p>Set Theory & Algebrahttp://gateoverflow.in/121592/set-theory-and-algebra?show=121612#a121612Tue, 14 Mar 2017 18:05:37 +0000Answered: set theory
http://gateoverflow.in/121590/set-theory?show=121602#a121602
<p>In order for a function <em>f</em>: <em>X</em> → <em>Y</em> to have an inverse, it must have the property that for every <em>y</em> in <em>Y</em> there must be one, and only one <em>x</em> in <em>X</em> so that <em>f</em>(<em>x</em>) = <em>y</em>. This property ensures that a function <em>g</em>: <em>Y</em> → <em>X</em> will exist having the necessary relationship with <em>f</em>.</p>
<p>so,here,domain is all real numbers and range is only positive real numbers.</p>
<p>so,-1 and 1 will map to 1 only.similarly -2 and 2 will map to 2 only,</p>
<p>so,here for y=1 ,we will get 2 values (1 and -1).and similarly for others.</p>
<p>and this function is not invertible.</p>
<p>but if we restrict our domain to positive real numbers only then this function would be invertible.</p>Set Theory & Algebrahttp://gateoverflow.in/121590/set-theory?show=121602#a121602Tue, 14 Mar 2017 17:12:34 +0000set theory
http://gateoverflow.in/121589/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11878804339871992571"></p>Set Theory & Algebrahttp://gateoverflow.in/121589/set-theoryTue, 14 Mar 2017 15:13:55 +0000set theory
http://gateoverflow.in/121586/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11342950768566882571"></p>Set Theory & Algebrahttp://gateoverflow.in/121586/set-theoryTue, 14 Mar 2017 14:58:39 +0000Answered: set theory
http://gateoverflow.in/121583/set-theory?show=121585#a121585
A) one-to-one & onto and so bijection.<br />
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B) not one-to-one because x=1 and x = -1 will have same f(x) value. not onto because 0 and negative numbers in the co-domain dont have pre-image. EX: -1 in co-domain need sqrt(-1) as pre-image but it is a complex number,which is not in the domain.<br />
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C) one-to-one & onto and so bijection.<br />
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D) not one-to-one because x = 1 and x = -1 will map to the same element in the co-domain. Also it is not onto because Negative numbers in the co-domain dont have any pre-image. Ex: -1 in co-domain needs a sqrt(-1.5) as its preimage which is a complex number and so not in our domain.Set Theory & Algebrahttp://gateoverflow.in/121583/set-theory?show=121585#a121585Tue, 14 Mar 2017 14:54:45 +0000set theory
http://gateoverflow.in/121581/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=6045207721206279828"></p>Set Theory & Algebrahttp://gateoverflow.in/121581/set-theoryTue, 14 Mar 2017 14:18:04 +0000Answered: set theory
http://gateoverflow.in/121574/set-theory?show=121576#a121576
<p>A function is called as ONTO if every element in the co-domain has a pre-image in the domain.</p>
<p>A) <em><strong>It is an ONTO function.</strong></em></p>
<p>Every even number in the co-domain can be made to have a pre-image by giving n = 0 and appropriate values for m. Ex: m=1,n=0 will have f(m,n) = 2 . similiarly,m=2,n=0 f(m,n) = 4 .....so on.</p>
<p>Every odd number in the co-domain can be made to have a pre-domain by giving n = 1 and appropriate values for m. Ex: m =1 ,n =1 will give f(m,n) = 1. m =2 ,n =1 will give f(m,n) = 3 ...and so on.</p>
<p>similairly we can also find pre-image for negative numbers in the codomain. Ex : m = 0,n =1 will give f(m,n) = -1 and so on.</p>
<p>So whatever element in the codomain we will be able to find a pre-image.so this function is ONTO.</p>
<p>B) <em><strong>It is not ONTO function</strong></em> as 6 in the co-domain will not have a pre-image.(no two square numbers are at a distance of 6....as 3<sup>2</sup>-2<sup>2</sup> = 5 , 4<sup>2</sup>-3<sup>2</sup> = 7 , 5<sup>2</sup>-4<sup>2</sup> = 9 .....)</p>
<p>C)<em><strong> it is an ONTO function</strong></em> as we can give n = -1 and values for m to get the co-domain elements. ex : if we want to find the pre-image for 1,then n= -1 and m = 1. If we want pre-image for 2, then n =-1 and m = 2 ... and so on.</p>
<p>D) <em><strong>It is an ONTO function</strong></em> as we can give values n = 0 and appropriate values for m to get all positive numbers and m = 0 and appropriate values for n to get all negative numbers. so any given number in co-domain we can find a pre-image. ex : if we want to find pre-image for 2 in co-domain then m = 2 and n = 0 will map to 2 in codomain.</p>
<p>E) <em><strong>NOT AN ONTO FUNCTION</strong></em> as we dont have pre-image for 1 in co-domain.</p>Set Theory & Algebrahttp://gateoverflow.in/121574/set-theory?show=121576#a121576Tue, 14 Mar 2017 13:26:46 +0000Answered: set theory
http://gateoverflow.in/121441/set-theory?show=121538#a121538
26.<br />
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so we are given A is subset of C and B is subset of D and we have to prove A$A\times B is subset of C\times D$<br />
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now to prove this let us assume we have a pair (x,y), where x belongs to A and y belongs to B<br />
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so we have to established that (x,y) belongs to A\times B so to establish this let us take an ordered pair (x,y) such that we can write it as x belongs to A and y belongs to B . now as x\epsilon A therefore it will belong to C as A is subset of C<br />
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and same goes for y<br />
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$so (x,y) \varepsilon C\times D$<br />
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as (x,y) is any ordered pair which belong to $A\varepsilon B$ and we have proved that it also belongs to $C\varepsilon D$<br />
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for more clarlity.. watch trev tutor videos on youtube for direct proofSet Theory & Algebrahttp://gateoverflow.in/121441/set-theory?show=121538#a121538Tue, 14 Mar 2017 05:30:03 +0000Answered: set theory
http://gateoverflow.in/121438/set-theory?show=121440#a121440
power set of a set is the set of all subsets including itself and empty set<br />
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So answer is DSet Theory & Algebrahttp://gateoverflow.in/121438/set-theory?show=121440#a121440Mon, 13 Mar 2017 05:30:59 +0000set theory
http://gateoverflow.in/121433/set-theory
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=10183339641406711391"></p>Set Theory & Algebrahttp://gateoverflow.in/121433/set-theoryMon, 13 Mar 2017 05:04:43 +0000Answered: set theory
http://gateoverflow.in/121428/set-theory?show=121432#a121432
A={2};<br />
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B = { { 1,2 }, 2 , {2} ,{ { 2 } } }Set Theory & Algebrahttp://gateoverflow.in/121428/set-theory?show=121432#a121432Mon, 13 Mar 2017 05:02:20 +0000