TL;DR

The text claims that this is an example of "proof by contradiction", but it seems rather to be a "proof by contrapositive."

I think that this is rather an example of "proof by contrapositive": given a rational ##a##, instead of proving directly that
##b \text{ irrational} \Rightarrow ab \text{ irrational}##
they have proved the contraposition,
##ab \text{ rational} \Rightarrow b \text{ rational}##.

They did not use in their proof an assumption of ##b## being irrational. Thus, they did not produce any contradiction.

TL;DR: The text claims that this is an example of "proof by contradiction", but it seems rather to be a "proof by contrapositive."

This is the example in question:

View attachment 371514

I think that this is rather an example of "proof by contrapositive": given a rational ##a##, instead of proving directly that
##b \text{ irrational} \Rightarrow ab \text{ irrational}##
they have proved the contraposition,
##ab \text{ rational} \Rightarrow b \text{ rational}##.

They did not use in their proof an assumption of ##b## being irrational. Thus, they did not produce any contradiction.

they have proved the contraposition,
##ab \text{ rational} \Rightarrow b \text{ rational}##.

They did not use in their proof an assumption of ##b## being irrational. Thus, they did not produce any contradiction.

"Suppose 𝑎 is a positive real number. If 𝑎 is irrational, then √𝑎 is irrational."

The proof:

Suppose to the contrary that the statement fails. That is, suppose that ##\sqrt{a} \in \mathbb{Q}##.
Then ##\exists m,n \in \mathbb{Z}, n \neq 0, \sqrt{a}=\frac m n##.
Now ##a=(\sqrt{a})^2=\frac {m^2} {n^2}## where ##m^2,n^2 \in \mathbb{Z}, n^2 \neq 0##.
That is ##a \in \mathbb{Q}##, which contradicts the condition that ##a## is irrational.

Is this a proof by contradiction or by contrapositive?

I have the same question about the following proof of a different statement:

"Suppose 𝑎 is a positive real number. If 𝑎 is irrational, then √𝑎 is irrational."

The proof:

Suppose to the contrary that the statement fails. That is, suppose that ##\sqrt{a} \in \mathbb{Q}##.
Then ##\exists m,n \in \mathbb{Z}, n \neq 0, \sqrt{a}=\frac m n##.
Now ##a=(\sqrt{a})^2=\frac {m^2} {n^2}## where ##m^2,n^2 \in \mathbb{Z}, n^2 \neq 0##.
That is ##a \in \mathbb{Q}##, which contradicts the condition that ##a## is irrational.

Is this a proof by contradiction or by contrapositive?

What the proof you have does is instead assume the statement is false

I have the same question about the following proof of a different statement:

"Suppose 𝑎 is a positive real number. If 𝑎 is irrational, then √𝑎 is irrational."

The proof:

Suppose to the contrary that the statement fails. That is, suppose that ##\sqrt{a} \in \mathbb{Q}##.
Then ##\exists m,n \in \mathbb{Z}, n \neq 0, \sqrt{a}=\frac m n##.
Now ##a=(\sqrt{a})^2=\frac {m^2} {n^2}## where ##m^2,n^2 \in \mathbb{Z}, n^2 \neq 0##.
That is ##a \in \mathbb{Q}##, which contradicts the condition that ##a## is irrational.

Is this a proof by contradiction or by contrapositive?

A proof by contradiction could lead to a different contradiction, unrelated to the original hypothesis. In this case, the proof cannot be immediately rewritten using contraposotion.

There's a discussion here:

https://math.stackexchange.com/ques...-contradiction-vs-proof-of-the-contrapositive