Answer to Question #203261 in Real Analysis for Raj kumar

Question

Are the following statements true or false? Give reasons for tour answers.

a) −2 isalimitpointoftheinterval ]−3,2].

b) The series (1/2) - (1/6) + (1/10) (−1/4) +.... is divergent.

c) The function, f (x) = sin²x is uniformly continuous in the interval [0,π].

d) Every continuous function is differentiable.

e) The function f defined on R by

f(x)= {0, if x is rational and 2, if x is irrational

Is integral element in the interval [2,3].

Answer 1

a) true

Notation $(a,b]$ means that $b$ is included in the interval.

b) false

Since $\frac{|a_{n+1}|}{|a_n|}<1$ , the series is convergent.

c) false

 The function f(x) is uniformly continuous in the interval $[0,2\pi]$ if

$\forall\ \varepsilon\ \exists\ \delta>0\ \forall\ x_1,x_1\isin[0,2\pi]\ (|x_1-x_2|<\delta)\implies (f(x_1)-f(x_2)<\varepsilon)$

$Let\ \varepsilon=1$ and $sin^2x_1=0,sin^2x_2=\pm 1$

Then:

$x_1^2=\pi n,x_2=\pi n+\pi/2$

$|x_1-x_2=\sqrt{\pi n+\pi/2}-\sqrt{\pi n}=\frac{\pi/2}{\sqrt{\pi n+\pi/2}+\sqrt{\pi n}}<\frac{2}{2\sqrt{\pi n}}<\frac{1}{\sqrt{n}}$

If $n>1/\delta^2$ then $|x_1-x_2|<\delta$ , but$|f(x_1)-f(x_2)|=1$

So, the given function is not uniformly continuous in the interval [0,π].

d) false

For example, continuous function $f(x)=|x|$ is not differentiable at x=0

e) false

An element b of a commutative ring B is said to be integral over A, a subring of B, if there are n ≥ 1 and a_j in A such that

$b^n+a_{n-1}b^{n-1}+...+a_1b+a_0=0$

For the given function:

 if x is irrational, then there are no coefficients $x_0,x_1,...,x_n$ , such that:

$2^n+2^{n-1}\cdot x_{n-1}+...+2x_1+x_0=0$