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

[Math Amateur]

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

 

[HallsofIvy]

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.

 

[Math Amateur]

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 \(\displaystyle a \in \mathbb{Q}\) and \(\displaystyle b \in \mathbb{R}\) \ \(\displaystyle \mathbb{Q}\) be any non-zero numbers ...

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

\(\displaystyle a \cdot b = c \Longrightarrow b = a^{ -1} \cdot c\)

But ... given that \(\displaystyle a\) is rational, we have that \(\displaystyle a^{ -1 }\) is rational ... ...

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

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

Peter

 

[HallsofIvy]

Yes, assuming that you have already proved that the product of two rational numbers is rational, that is a valid proof.

 

[Math Amateur]

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