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I Need to prove this by contradiction :

If $a$ is Irrational then $\frac{2a-3}{2a+3}$ is Irrational.

I did: Iff $p$ is Irrational, then $\frac{2a-3}{2a+3}$ is Rational and a Rational number can be written as $\frac{p}{q}$ where p and q are integers and q is different than zero so,

$\dfrac{2a-3}{2a+3}=\dfrac{p}{q}$ which is

$$q(2a-3)=p(2a+3)=$$

$$2aq-3q=2ap+3p=$$

$$2aq-2ap=3p+3q=$$

$$a=\frac{3p+3q}{2q-2p},$$ therefore $a$ is rational and $2q-2p$ cannot be zero. So I proved it by contradiction.

Am I doing it right? can anyone help me?

Thanks

Michael Hardy
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user290335
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  • Looks good,but better to set up as proof by contradiction, i.e. start by assuming $(2a-3)/(2a+3)$ rational, then go on as you did. – coffeemath Feb 08 '16 at 16:47
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    Essentially it's right, but $(1)$ You shouldn't start by saying "Iff $p$ is rational", but rather by assuming $(2a-3)/(2a+3)$ is rational, and $(2)$ you should write "$=$" only when you're saying two things are equal. Thus $2aq-3q=2ap+3p$ is right, but you should have the next "equals" sign saying that's equal to what you have on the next line. $\qquad$ – Michael Hardy Feb 08 '16 at 17:16

2 Answers2

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That's the right idea, but you should be more clear that you're setting up a proof by contradiction. There are also two other issues:

  • You have a typo on the last line: the numerator should be $3(p+q)$
  • How can you be sure that $2(q - p) \neq 0$? You should state assumptions about $p$ and $q$ initially that ensure they are not equal. (To do this you need to rule out $(2a-3)/(2a+3) = 1$, but that's not difficult.)
Dan
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  • So to be correct I need to prove that 2(q-p) is different than zero? – user290335 Feb 08 '16 at 16:51
  • If you don't know that $2(q-p) \neq 0$, then you can't be sure that you can divide by it in your last step. – Dan Feb 08 '16 at 16:52
  • You could also avoid this by instead just noting that $2(q-p)a = 3(p+q)$ is an equation of the form $ax=y$ for rational numbers $x$ and $y$. Hence either $x = 0$ or $a$ is rational, but then you still have to address the case where $x = 2(q-p) = 0$. – Dan Feb 08 '16 at 16:54
  • Yes I understood. How can I do that? Can I state it by assuming it or I do I need to find a proof? – user290335 Feb 08 '16 at 16:54
  • Well, if $(2a-3)/(2a+3) = 1$, then $2a-3 = 2a+3$ and hence $6 = 0$, so you can simply note at the beginning that $p \neq q$ because of this. – Dan Feb 08 '16 at 16:55
  • I did not understand sorry can you explain it better? – user290335 Feb 08 '16 at 17:04
  • If $2(p-q) = 0$ then $p=q$. If $p=q$, then (from the definition of $p$ and $q$) you've assumed that $\frac{2a-3}{2a+3} = \frac{p}{q} = 1$. The equation $\frac{2a-3}{2a+3} = 1$ can be simplified to $6 = 0$, which is impossible. Therefore $p \neq q$. – Dan Feb 08 '16 at 17:06
  • Ah ok, it makes sense, thank you for your help. – user290335 Feb 08 '16 at 17:10
  • Sure thing! Glad it's making sense. :) – Dan Feb 08 '16 at 17:10
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I think you can be a little clearer.

Let $a$ be irrational. Say $\dfrac{2a-2}{2a+3}$ is rational. Then, $\dfrac{2a-2}{2a+3}=\dfrac{p}{q}$, where $p,q\epsilon \Bbb{Z}$ and $q$ is not zero.

Show why $p$ cannot equal $q$, then show that $a$ is rational, which is a contradiction. Hence, if $a$ is irrational, then $\dfrac{2a-2}{2a+3}$ must also be rational.

settheorynoob
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