If the square of an irrational number r is irrational, can it be equal to a + br, where both a and b are rational

Question

I am trying to evaluate such a statement:

$$\forall r \in \mathbb{R} \setminus \mathbb{Q}: \ ({r^{2}} \notin \mathbb{Q} \implies \forall a \in \mathbb{Q} \ \forall b \in \mathbb{Q}: {r^{2}} \neq a + br)$$

but it seems to exceed my skills. Could you please help?

Consider the contrapositive. Then it would be easy to find a counterexample. — player3236, Nov 10 '20 at 04:16

Derek Luna · Accepted Answer · 2020-11-10T06:10:21.597

4

The following is a counterexample:

Set $r = \sqrt{2} +1 \in \mathbb{I}$.

Then $r^{2} = (\sqrt{2} +1)^2=1+2\sqrt{2}+2=1+2(\sqrt{2}+1) = 1+2r.$

edited Nov 10 '20 at 06:10

answered Nov 10 '20 at 04:37

Derek Luna

2,732
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Thank you! So simple! I was convinced that the statement was true so I spent so much effort looking for a proof instead of investigating these kinds of examples. – Arkadiusz Filipczyk Nov 10 '20 at 06:09
You're welcome! – Derek Luna Nov 10 '20 at 06:09

If the square of an irrational number r is irrational, can it be equal to a + br, where both a and b are rational

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