General Mathematics

Question Solution : we discuss options one by one. Recall that a point $p\in\mathbb{R}$ is said to be an interior point of $S\subseteq\mathbb{R}$ if  i) $p\in S$ ii) there exists a neighborhood of $p$ which is contained in $S$. Now given that, $2018\in S$ is an interior point of $S$. Hence by definition there exists a neighborhood of $2018$ which is contained in $S$, so that $S$ contains an interval. Hence (A) is true. Now as $S\subseteq\mathbb{R}$, and $2018\in S$ is interior point of $S$ hence, it is an limit point of $S$ and hence there is sequence of points in $S$ which do not converge to $2018$. Hence (B) is also true. Clearly (C) is also true by (A) (since $S$ contains a neighborhood (interval) containing $2018$ and we know by definition every point inside that interval is also an interior point of $S$ ) Finally for (D): we know by definition of an interior point $S$ contains a neighborhood (which is an interval) of point $2018$. Now this neighborhood can be too small such...

General Mathematics Question : Given $(x_n)$ is convergent sequence in $\mathbb{R}$ and $(y_n)$ is bounded sequence in $\mathbb{R}$ then we can conclude that, (a) $(x_n+y_n)$ is convergent. (b) $(x_n+y_n)$ is bounded. (c) $(x_n+y_n)$ has no convergent subsequence. (d)$(x_n+y_n)$ has no bounded subsequence. Solution : we discuss options one by one (a) take $x_n=1$ for all $n\in\mathbb{N}$ and $y_n=(-1)^n$ for all $n\in\mathbb{N}$ then, sequence $(x_n)$ is convergent and $(y_n)$ is bounded. But, $(x_n+y_n)=(0,2,0,2,0,2,....)$ which is not convergent. Hence (a) is false. (b) given that sequence $(x_n)$ is convergent and hence it is bounded. Also given that, $(y_n)$ is bounded. So that sum of two bounded sequence $(x_n+y_n)$ is also bounded. Hence (b) is true. (c), (d): By option (b) as sequence $(x_n+y_n)$ is bounded sequence and hence by "Bolzano Weierstrass theorem for sequences", it has convergent (hence bounded) subsequence.  So that, (c),(d) are fa...

General Mathematics Question Solution : we know that,  Number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is $gcd(m,n)$ Now, as $221=13\times 17$. Hence, $gcd(51,221)≠1$, $gcd(91,221)≠1$, $gcd(119,221)≠1$, But $gcd(21,221)=1$ Hence number of homomorphisms from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$ is $1$ that is, there is only one homomorphism from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$. So that this must be trivial homomorphism( zero map).  Hence if $f:\mathbb{Z}_{21}\rightarrow\mathbb{Z}_{221}$ is homomorphism then it is trivial homomorphism ( zero map) i.e. $f(a)=0$ for all $a\in\mathbb{Z}_{21}$. Hence option (A) is correct.  (Note other options are not correct because in those case there exists non-trivial homomorphisms, because gcd of numbers $51,91,119$ with $221$ is not $1$.) Hence answer is "option (A)" only. Book recommend : see below, Must buy it. Nice offer must vist! 

General Mathematics Question Solution: we discuss options one by one: (A) if $P$, and $Q$ are compact subsets of $\mathbb{R}$ then $P$ and $Q$ both are bounded and closed subsets of $\mathbb{R}$ (By "Heine-Borel theorem"). Also we know that,  Union of two bounded subsets of $\mathbb{R}$ is bounded subset of $\mathbb{R}$ Finite union of closed sets in $\mathbb{R}$ is again closed set in $\mathbb{R}$. Hence, $P\cup Q$ is again bounded and closed subset of $\mathbb{R}$ and hence by "Heine Borel theorem" it is compact. So that option (A) is true. (B) take $P=\mathbb{Q}$ and $\mathbb{Q^c}$ then both $P$ and $Q$ are nonempty disjoint subsets of $\mathbb{R}$ that are not connected, but their union is $\mathbb{R}=(-∞,+∞)$ is connected subset of $\mathbb{R}$. Hence (B) is false.  (C) take $P=\{1,2\}$, $Q=(1,2)$ then $P$ is closed, and $P\cup Q=[1,2]$ is closed. Also both $P,Q$ are nonempty disjoint subsets of $\mathbb{R}$. But, $Q$ is not closed subset of $\m...

General Mathematics Question :  Solution : Note that,   An element $m$ is maximum of set $S$ iff $s≤m$ for all $s\in S$ and $m\in S$. Further, if a set $S$ has maximum element then it is supremum of $S$. Similarly, an element $l$ is minimum or least element of set $S$ if $l≤s$ all $s\in S$ and $l\in S$. Further, if a set $S$ has minimum or least element then it is infimum of $S$. Hence, $sup\{a_n: n\in\mathbb{N}\}$ $=sup\{a_1,a_2,a_3,...\}$ $=sup\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$ $=3$ (since maximum  of set $\{a_1,a_2,...\}$ is 3.   ) Similarly,  $inf\{a_n: n\in\mathbb{N}\}$ $=inf\{a_1,a_2,a_3,...\}$ $=inf\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$ $=1$ (since, $\{a_1,a_2,...\}$ has minimum or least element 1)  Hence option (A) is true, (C) is false.  Now to discuss options (B) and (C): we note that,  $\text{lim inf} (a_n)$ is infimum of set of all subsequential limits of sequence $(a_n)$ and similarly, $\te...

General Mathematics Question : Let $G$ be a non abelian group, $y\in G$. let the maps $f, g, h$ from $G$ to itself is defined by  $$f(x)=yxy^{-1}, g(x)=x^{-1}\text{ and }h=g•g$$ Then, (A) $g$ and $h$ are homomorphism and $f$  is not homomorphism. (B) $h$ is homomorphism, $g$ is not homomorphism. (C) $f$ is homomorphism, $g$ is not homomorphism. (D) $f,g,h$ all are homomorphism. Solution :  Since $y^{-1}y=e\in G$ where $e$ is identity element in $G$. Hence we have, $f(ab)=yaby^{-1}=yay^{-1}yby^{-1}=(yay^{-1})(yby^{-1})=f(a)f(b)$ for all $a,b\in G$. Hence by definition of homomorphism $f$ is homomorphism.  Note that, $g$ is not a homomorphism since  $g(ab)=(ab)^{-1}=b^{-1}a^{-1}$. Whereas,  $g(a)g(b)=a^{-1}b^{-1}$ so that  $g(ab)$ not always equal to $g(a)g(b)$ ( since there are some $a,b\in G$ such that $b^{-1}a^{-1}≠a^{-1}b^{-1}$ as $G$ is non-abelian, otherwise $G$ will be abelian) Finally as,   $h(ab)=g•g(ab)=g(g(ab))=g(b^{-1...

General Mathematics Question: Let $G$ be non-cyclic group of order $4$. Consider the statements I and II. I. There is no injective (one-one) homomorphism from $G$ to $\mathbb{Z}_8$ II. There is no surjective (onto) homomorphism from $\mathbb{Z}_8$ to $G$ Then, (A) I is true          (B) I is false (C) II is true         (D) II is false Solution :   let suppose that $f:G\rightarrow \mathbb{Z}_8$ is an injective homomorphism. Then By Fundamental theorem of homomorphism (First isomorphism theorem ) we have,  $$\frac{G}{ker(f)}≈f(G)$$ But, $ker(f)=\{e\}$ (since $f$ is injective i.e. one-one map ) Hence above implies, $$G≈f(G)$$ But $f(G)$ is subgroup of $\mathbb{Z}_8$ and hence must be cyclic(Since, subgroup of a cyclic group is cyclic) So that $G$ is cyclic. Contradiction to fact that $G$ is non-cyclic group and hence our assumption must be wrong! Hence, there is No injective (one-one) homomorphism fr...