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I'm trying to prove that not all irrational numbers are of the form c=a+b√2 when a,b are rational. I'm thinking if I prove that π does not have this form, them there's an infinity of irrationals which can be written like c=a+bπ with a,b rationals cannot be written in the other form. I'm trying to prove it through contradiction. Therefore I'm supposing π=a+b√2. Then π-b√2=a. Now we have two possibilities. If a≠0 then a is irrational through the properties of irrational numbers which is therefore a contradiction. If a=0 then that means π=b√2. Now, I imagine the next step would lead to showing that b is an irrational number. However I do not know how to proceed to show this is the case. Thanks in advance.

Jyrki Lahtonen
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Miguel Angel Andrade Velázquez
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    The short answer is that you picked the wrong example. Proving that $\pi$ is not of this form is somewhat taxing, and requires deeper theory. Try $\sqrt3$ instead. – Jyrki Lahtonen Feb 24 '20 at 04:55
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    Also, your logic leaks. There are no general properties of irrational numbers that would justify your claim: If a≠0 then a is irrational through the properties of irrational numbers which is therefore a contradiction. – Jyrki Lahtonen Feb 24 '20 at 04:57
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    Mind you, it is true that $\pi- b\sqrt2$ is irrational for all rational $b$, but A) that's exactly the deeper stuff, B) and is equivalent to your claim (making the argument circular). My advice is to forget about $\pi$. – Jyrki Lahtonen Feb 24 '20 at 05:00
  • Thanks a lot, the proof with \sqrt3 was way easier – Miguel Angel Andrade Velázquez Feb 24 '20 at 05:55

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