GATE Overflow - Recent questions and answers in Set Theory & Algebra
http://gateoverflow.in/qa/mathematics/discrete-mathematics/set-theory-%26-algebra
Powered by Question2AnswerAnswered: 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 02:13:39 +0000Answered: GATE2005-8
http://gateoverflow.in/1157/gate2005-8?show=126515#a126515
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=3536398992072172462"></p>
<p>From the venn diagram it is clear that, <strong>X=Y (Option A)</strong></p>Set Theory & Algebrahttp://gateoverflow.in/1157/gate2005-8?show=126515#a126515Fri, 21 Apr 2017 23:59:29 +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 15:28:05 +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 12:19:10 +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 12:09:18 +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 04:08:56 +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 15:35:58 +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 15:15:09 +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 14:58:05 +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 13:55:02 +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 12:50:17 +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 12:28:12 +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 00:47:33 +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 01:00:40 +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 00:48:28 +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 05:27:57 +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 18:26:54 +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 15:56:06 +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 12:59:27 +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 10:09:43 +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#a124106Tue, 04 Apr 2017 19:16:56 +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 17:47:47 +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 15:22:27 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123963/isi-2004-miii?show=124008#a124008
<p>$$\begin{align*} &\Rightarrow \frac{2-gx+x^{2}}{1-x+x^{2}}\leq 3 \\ &\Rightarrow 2-gx+x^{2} \leq 3 - 3x + 3x^2 \\ &\Rightarrow 2x^2 + \left ( g-3 \right )x + 1 \geq 0 \\ \end{align*}$$</p>
<p>So root will be imaginary and the curve never touches the $X$ axis. Like the following curve :</p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11966087500766096828"></p>
<p>$$\begin{align*} &\Rightarrow B^2 - 4AC < 0 \\ &\Rightarrow \left ( g-3 \right )^2 - 4.2.1 <0 \\ &\Rightarrow -\sqrt{8} < \left ( g-3 \right ) < +\sqrt{8} \\ &\Rightarrow -\sqrt{8}+3 < g < +\sqrt{8}+3 \\ &\Rightarrow -5.83 < g < +5.83 \\ &\Rightarrow \text{Option B:} \quad {\bf \color{green}{-1 \leq g \leq +1}} \qquad \left [ \text{well inside range} \right ] \end{align*}$$</p>
<p><strong>(B) is the answer.</strong></p>Set Theory & Algebrahttp://gateoverflow.in/123963/isi-2004-miii?show=124008#a124008Tue, 04 Apr 2017 12:05:59 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123882/isi-2004-miii?show=123906#a123906
<p>Given ,</p>
<p>α1, α2,.. .αn are the positive numbers , </p>
<p>So (α1 / α2) , ( α2 / α3 ) etc will be also positive numbers..</p>
<p>Hence applying A.M. >= G.M ,</p>
<p> ( (α1 / α2) + ( α2 / α3 ) + ............ + ( αn / α1 ) ) / n >= ( (α1 / α2) . ( α2 / α3 ) ............... ( αn / α1 ) )<sup>1/n</sup></p>
<p>==> ( (α1 / α2) + ( α2 / α3 ) + ............ + ( αn / α1 ) ) >= n . ( (α1 / α2) . ( α2 / α3 ) ............... ( αn / α1 ) )<sup>1/n</sup></p>
<p>==> ( (α1 / α2) + ( α2 / α3 ) + ............ + ( αn / α1 ) ) >= n . 1<sup>1/n </sup></p>
<p>==> ( (α1 / α2) + ( α2 / α3 ) + ............ + ( αn / α1 ) ) >= n</p>
<p><em><strong>Hence A) is the correct answer..</strong></em></p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/123882/isi-2004-miii?show=123906#a123906Tue, 04 Apr 2017 05:25:09 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123818/isi-2004-miii?show=123826#a123826
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=3091194470720553436"></p>
<p>c is ans.</p>Set Theory & Algebrahttp://gateoverflow.in/123818/isi-2004-miii?show=123826#a123826Mon, 03 Apr 2017 17:40:33 +0000Answered: ISI 2004 MIII
http://gateoverflow.in/123813/isi-2004-miii?show=123822#a123822
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11296738527660479016"></p>
<p>so ans is 2.</p>Set Theory & Algebrahttp://gateoverflow.in/123813/isi-2004-miii?show=123822#a123822Mon, 03 Apr 2017 17:29:07 +0000Answered: GATE2014-2-5
http://gateoverflow.in/1957/gate2014-2-5?show=123737#a123737
<h3>The graph of a degree 3 polynomial f(x) = a0 + a1x + a2(x^2) + a3(x^3), where a3 ≠ 0 is a cubic curve, as can be seen here <a rel="nofollow" href="https://en.wikipedia.org/wiki/Polynomial#/media/File:Polynomialdeg3.svg">https://en.wikipedia.org/wiki/...</a> Now as given, the polynomial is zero at x = 1, x = 2 and x = 3, i.e. these are the only 3 real roots of this polynomial. Hence we can write the polynomial as f(x) = K (x-1)(x-2)(x-3) where K is some constant coefficient. Now f(0) = -6K and f(4) = 6K ( by putting x = 0 and x = 4 in the above polynomial ) and f(0)*f(4) = -36(k^2), which is always negative. Hence option A. We can also get the answer by just looking at the graph. At x < 1, the cubic graph (or say f(x) ) is at one side of x-axis, and at x > 3 it should be at other side of x-axis. Hence +ve and -ve values, whose multiplication gives negative.</h3>Set Theory & Algebrahttp://gateoverflow.in/1957/gate2014-2-5?show=123737#a123737Mon, 03 Apr 2017 09:46:50 +0000Answered: TIFR-2011-Maths-A-16
http://gateoverflow.in/30197/tifr-2011-maths-a-16?show=123696#a123696
<p>I THINK ANS IS NO...</p>
<p><em>x</em><sup>4</sup>+7<em>x</em><sup>3</sup>−13<em>x</em><sup>2</sup>+11<em>x</em></p>
<p>=>x(x<sup>3</sup>+7x<sup>2</sup>-13x+11)</p>
<p>so '0' is a root of the polynomial....</p>
<p>Again the coefficient of the given polynomial are real ... so if there are complex root ,they are two in number...,</p>Set Theory & Algebrahttp://gateoverflow.in/30197/tifr-2011-maths-a-16?show=123696#a123696Mon, 03 Apr 2017 06:23:45 +0000Answered: Rosen(relations)
http://gateoverflow.in/123319/rosen-relations?show=123471#a123471
<p>By changing direction of edges, self loop will remain as it is as well as isolated vertex.</p>
<p>Eg: R={(1,1),(1,2)}</p>
<p>R<sup>-1</sup>={(1,1),(2,1)}</p>
<p>In R there was an edge starting form vertex 1 and ending at 2</p>
<p>In R<sup>-1</sup> there is an edge starting from 2 and ending at 1</p>Set Theory & Algebrahttp://gateoverflow.in/123319/rosen-relations?show=123471#a123471Sun, 02 Apr 2017 04:44:29 +0000Answered: ISI 2017
http://gateoverflow.in/123274/isi-2017?show=123345#a123345
i will write 'a' for alpha<br />
<br />
$x(a-x) < y(a-y) \\ ax-x^2 -ay+y^2<0 \\ a(x-y)-(x-y)(x+y)<0 \\ (x-y)(a-(x+y))<0 \\ x<y \implies a>(x+y) \\ x+y<2 \implies a\geq 2$Set Theory & Algebrahttp://gateoverflow.in/123274/isi-2017?show=123345#a123345Sat, 01 Apr 2017 15:26:48 +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 14:08:35 +0000Answered: GATE2017-1-47
http://gateoverflow.in/118330/gate2017-1-47?show=123284#a123284
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=16646818402717939195"></p>
<p>solution</p>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/118330/gate2017-1-47?show=123284#a123284Sat, 01 Apr 2017 10:01:06 +0000Answered: GATE2014-3-49
http://gateoverflow.in/2083/gate2014-3-49?show=123239#a123239
This kind of functions are called identity functions. We assume f(i) = k. So, f(k) = i. Now, since the values of ' i ' and ' j ' would be same for atleast some values if the domain and co - domain intersect, which is true for the given question, Q is definitely true. But this might not happen for all the values of ' i ', hence, P is not always true. Now, ' i ' ranges from 0 to 2014, so, it takes 2015 possible values. From the definition of a function, we know that for each input to the function, we have a unique output. Also, in the given question, domain and co - domain are exactly same. Therefore, the function is onto and hence, R is definitely true. Thus, the correct option is B.Set Theory & Algebrahttp://gateoverflow.in/2083/gate2014-3-49?show=123239#a123239Sat, 01 Apr 2017 06:07:15 +0000Answered: Rosen (relation)
http://gateoverflow.in/123125/rosen-relation?show=123135#a123135
<p>Since there are 4 elements in the set, so the total possible relations on set is $2^{2^{4}}=2^{16}$</p>
<p>There are total 16 possible ordered pairs on the set.</p>
<p>Now since it is given that element (a,a) should be present in all the relations. So of the remaining 15 ordered pairs any of them can be either part of the relation or not.</p>
<p>$\binom{15}{0} + \binom{15}{1} + ... + \binom{15}{15}= 2^{15}$ </p>
<p>So ans is 2<sup>15</sup></p>Set Theory & Algebrahttp://gateoverflow.in/123125/rosen-relation?show=123135#a123135Fri, 31 Mar 2017 13:48:33 +0000Answered: GATE2014-3-16
http://gateoverflow.in/2050/gate2014-3-16?show=123050#a123050
<p><strong>Theorem- S is a countably infinite set, 2 S (the power set) is uncountably infinite. </strong></p>
<p><strong>Proof:</strong> We show 2 S is uncountably infinite by showing that 2 N is uncountably infinite. (Given the natural bijection that exists between 2 N and 2 S –because of the bijection that exists from N to S– it is sufficient to show that 2 N is uncountably infinite.) Assume that the set 2 N is countably infinite. The subsets of N can be listed A0, A1, A2, . . . so that every subset is Ai for some i. We define another set A = {i|i ≥ 0 and i 6∈ Ai} which contains those integers i which are not members of their namesake set Ai . But A is a subset of N, and so A = Aj for some j. But this means 1. If j ∈ A, then j 6∈ A. 2. If j 6∈ A, then j ∈ A. We have a contradiction, since j must either be in the set A or not in the set. Therefore 2 N is not countably infinite. ⋄ The Theorem that the power set of a countably infinite set is an uncountable set indicates that the set of all languages over any alphabet Σ, |Σ| 6= 0 is uncountable. (2 Σ∗ is an uncountable set.)</p>Set Theory & Algebrahttp://gateoverflow.in/2050/gate2014-3-16?show=123050#a123050Thu, 30 Mar 2017 19:37:26 +0000Answered: Suppose 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?show=122669#a122669
<p><strong>a b c</strong> are in<strong> GP</strong> i.e.,<strong> b<sup>2 </sup>=ac.... (GP property)</strong></p>
<p><strong>solve by example..</strong> let<strong> <em>a b c</em> </strong>be<strong> 2,4,8 respectively </strong>and it satisfies <strong>4<sup>2</sup>=2*8</strong>=16 (a, b, c > 0 {given} )</p>
<p>or,2<sup><strong>p</strong></sup> = 4<sup>q</sup> = 8 <sup>r </sup></p>
<p>and should also satisfy <strong>a<sup>p</sup> = b<sup>q</sup> = c<sup>r</sup> != 1 (given)</strong></p>
<p>if p=q=r=0 then a<sup>p</sup> = b<sup>q</sup> = c<sup>r</sup><strong> = 1 (option d false)</strong></p>
<p>let us take.... for above example 2<sup><strong>p</strong></sup> = 4<sup>q</sup> = 8 <sup>r </sup><strong>=64 ...so we get p=6,q=3,r=2</strong></p>
<p>therefore,</p>
<ul>
<li>pqr are not in GP <strong>(option A false)</strong></li>
<li>and, pqr are not in AP <strong>(option B false)</strong></li>
<li>and, pqr are in HP <strong>(option C true-ANSWER)</strong></li>
</ul>
<p> </p>Set Theory & Algebrahttp://gateoverflow.in/122639/suppose-a-b-c-0-are-in-geometric-progression-and-a?show=122669#a122669Mon, 27 Mar 2017 06:48:36 +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/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 +0000