Question on Negation of Definition of limit of Sequence

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By Akash Patalwanshi August 05, 2020

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

"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 have to learn something then you need to work by yourself)

Task: You can comment counterexamples to discard options (b), (c), (d) in below comment box. I will verify them and reply you.

Answer

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IIT JAM 2019 Abstract Algebra: Question (2)

By Akash Patalwanshi June 21, 2020

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...

IIT-JAM 2018 Abstract Algebra Question (3)

By Akash Patalwanshi June 25, 2020

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!

Series Question IIT JAM 2018

By Akash Patalwanshi July 06, 2020

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...

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