MHB Struggling with Exercise 2.1.40 in Sohrab's Basic Real Analysis?

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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with Corollary 2.1.39/Exercise 2.1.40 ...

Corollary 2.1.39/Exercise 2.1.40 reads as follows:https://www.physicsforums.com/attachments/7090I have not been able to make a meaningful start on Exercise 2.1.40 despite Sohrab's hint ...

Can someone please help with Exercise 2.1.40 ... ?

Peter
 
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The main theorem, Theorem 2.1.38, to which this is a corollary, says that between any two real numbers, there is a rational number. Applying that to the two numbers \frac{x}{\sqrt{2}} and \frac{y}{\sqrt{2}}, there exist a rational number, s, such that \frac{x}{\sqrt{2}}< s< \frac{y}{\sqrt{2}}. Multiply each part by \sqrt{2}: x< s\sqrt{2}< y.

To finish, show that any (non-zero) rational number times an irrational number is irrational.
 
HallsofIvy said:
The main theorem, Theorem 2.1.38, to which this is a corollary, says that between any two real numbers, there is a rational number. Applying that to the two numbers \frac{x}{\sqrt{2}} and \frac{y}{\sqrt{2}}, there exist a rational number, s, such that \frac{x}{\sqrt{2}}< s< \frac{y}{\sqrt{2}}. Multiply each part by \sqrt{2}: x< s\sqrt{2}< y.

To finish, show that any (non-zero) rational number times an irrational number is irrational.
Thanks for the help, HallsofIvy ...

As you say, we need to show that any (non-zero) rational number times an irrational number results in an irrational number.

So let $$a \in \mathbb{Q}$$ and $$b \in \mathbb{R}$$ \ $$\mathbb{Q}$$ be any non-zero numbers ...

Consider $$a \cdot b = c$$ for some real number $$c $$ ...Now ... assume that $$c$$ is rational ... then we have ...

$$a \cdot b = c \Longrightarrow b = a^{ -1} \cdot c$$

But ... given that $$a$$ is rational, we have that $$a^{ -1 }$$ is rational ... ...

... and it follows, under the assumption that $$c$$ is rational, that $$a^{ -1} \cdot c$$ is rational ...

But then $$b$$ must be rational ... Contradiction! ...So $$c$$ must be irrational ... that is, the product of a rational number and an irrational number is an irrational number ...Is that correct?

Peter
 
Yes, assuming that you have already proved that the product of two rational numbers is rational, that is a valid proof.
 
HallsofIvy said:
Yes, assuming that you have already proved that the product of two rational numbers is rational, that is a valid proof.
Thanks for your help, HallsofIvy ...

Definitely assumed that product f two rationals is rational ... thanks for pointing that out ...

Peter
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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