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How can I prove that the sum of two irrational numbers is most likely irrational number?

Lucie
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    Let's focus on the first question: is more irratinal or rational numbers[?] The best I can make out of this is "are there more irrational numbers than rational numbers?", but I don't see how it is connected to the first sentence. Can you clarify please? – rschwieb Nov 23 '16 at 18:24
  • I think what OP is asking is if we pick two arbitrary irrational numbers, what is more likely, the sum being rational or irrational? – Anurag A Nov 23 '16 at 18:33
  • ok, for example a, b, c are irrational numbers and d is rational number. I know that a+b=c or a+b=d. And the question is: When i do the sum of two irrational numbers - the result of the sum is mostly irrational number or is it mostly rational number. And I'm sorry for my english, I don't come from english speaking country. – Lucie Nov 23 '16 at 18:34
  • If you fix one irrational $\alpha$ then there are uncountably many irrationals $\beta$ such that $\alpha + \beta$ is irrational and only countably many $\gamma$ such that $\alpha + \gamma \in \mathbb Q$. Is that the sort of thing you are after? – lulu Nov 23 '16 at 18:35
  • Yes, Anurag A, that is what I wanna know. – Lucie Nov 23 '16 at 18:36
  • How to define "more likely" when each option corresponds to infinitely many choices? – Did Nov 23 '16 at 18:40
  • Yes, lulu, but I need to prove it somehow, my teacher is strict. I don't know how to prove it. His tasks are crazy. I think it's almost impossible. – Lucie Nov 23 '16 at 18:43

2 Answers2

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Let $\mathbb{S}$ denote the set $\mathbb{R}\backslash\mathbb{Q}$, then:

  • $\forall{x\in\mathbb{S}}:x+(1-x)\in\mathbb{Q}$
  • $\forall{x\in\mathbb{S}}:x+(x-1)\not\in\mathbb{Q}$

Therefore, since $\mathbb{S}$ is uncountable:

  • The amount of rational results is uncountable
  • The amount of irrational results is uncountable
barak manos
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  • How can rational results be uncountable? – orion Nov 23 '16 at 18:41
  • @orion: There are (uncountably) many results for each rational value. – barak manos Nov 23 '16 at 18:41
  • @orion: For example: $1=\sqrt2+(1-\sqrt2)=\pi+(1-\pi)=\phi+(1-\phi)=e+(1-e)=\dots$. – barak manos Nov 23 '16 at 18:43
  • Ok, thanks, but what is more likely, the sum being rational or irrational? and how would you prove it? – Lucie Nov 23 '16 at 20:19
  • @Lucie: I think that since there is an uncountable amount of different ways to generate each one of them, the probabilities are identical. But I think that it generally requires proving $|\mathbb{R}|=|\mathbb{R}\times\mathbb{R}|$, and I'm not really sure how to prove this one (or even whether or not it is true to begin with). – barak manos Nov 23 '16 at 21:12
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Given a fixed number $z$, you can take each $q\in \mathbb Q$ and find the unique $y_q\in \mathbb R$ such that $z+y_q=q$. Since each $y_q$ is unique, clearly there are only countably many numbers with that can be found.

So out of the entire pool of $|\mathbb R|$ things that can go into the $y$ of "$z+y$", only countably many will yield a rational number, and the rest (uncountably many) will yield an irrational number.

rschwieb
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  • But z and y have to be irrational, not rational. Sorry, I don't understand. – Lucie Nov 23 '16 at 20:09
  • @Lucie I don't understand what problem you are describing in your last comment. $z$ can be an irrational number, everything still applies. There are still only countably many $y_q$ all of which are necessarily irrational. – rschwieb Nov 23 '16 at 20:27