General Mathematics

Question Press/click below hints, answer "buttons" for solution:   Do you really need hint? Give your 100% first! If still you have problem then check hint 2 Hint 1 "given that $\lim_{x \to +\infty} a_{n}= 0$. Hence by definition of limit of sequence we have, for given any $\epsilon >0$ there exists $K\in\mathbb{N}$ such that, $|a_n-0|<\epsilon$ i.e. given any $\epsilon >0$ there exists $K\in\mathbb{N}$ such that, $|a_n|<\epsilon$ "The question ask negation of above statement" Which is given by "there exists an $\epsilon>0$ such that for every $K\in\mathbb{N}$ there exists $n>K$ such that $|a_n|\geq\epsilon$ from this can you conclude the required answer ? Hint 2 By Hint2 we conclude that the answer is option (a) Can you discard the other options? yes! Find counterexample for discarding other options. Did you find any? ("I will not do spoon feeding" If You hav...

Solution of above question is given following YouTube video, "please use headphones for better audio quality"  thank you! 

   Solution is given(explained) in following YouTube video

General Mathematics Question :  Solution :   $\sum_{n=1}^\infty (-1)^{n+1}\frac{a_{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n+1+\frac{1}{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!(n+1)}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n-1)!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n+1)!}$ $=[1-\frac{1}{1!}+\frac{1}{2!}-...]+ [1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[\mathbf{1-1}+1-\frac{1}{2!}+\frac{1}{3!}-...]+[\mathbf{1-\frac{1}{1!}}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-(1-\frac{1}{1!}+\frac{1}{2!}-...)]+e^{-1}$ $=e^{-1}+[1-e^{-1}]+ e^{-1}$ $=e^{-1}+1$ Hence answer is option (D) Book recommended : Must purchase it nice offer!   For our visitors : Please Support us By...

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...