i was wondering if someone could check my proof $Q= \{a/b , c,d : a,c ∈ \mathbb Z , b,d ∈ N>0\}$ $a/b =x+y$

$a/b -y=x$

proof by contradiction.

Let $x-y$ is rational

$c/d = x-y$

sub $a/b -y = x$ in for $x$

$c/d = (a/b -y) - y$

$a/b - c/d = 2y$

This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.

there it can be said $x-y$ is irrational

  • $\begingroup$ What does $\frac ab $ and $\frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion? $\endgroup$ – Saucy O'Path Apr 8 '19 at 12:15
  • $\begingroup$ Definition of a rational number states that the denominator and numerator must be in their lowest forms $\endgroup$ – josh Apr 8 '19 at 12:18
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    $\begingroup$ You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $\x\pm y\in \mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms". $\endgroup$ – lulu Apr 8 '19 at 12:18
  • $\begingroup$ @josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $\operatorname{gcd}(a,b)\ne1$ either). $\endgroup$ – Saucy O'Path Apr 8 '19 at 12:28
  • $\begingroup$ What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction $\endgroup$ – josh Apr 8 '19 at 12:33

Yes, your proof seems right.

I like the following writing.

If $x-y\in\mathbb Q$ so $x-y+x+y=2x\in\mathbb Q$, which says $x\in\mathbb Q$, which is a contradiction.

Id est, $x-y\not\in\mathbb Q$.

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  • $\begingroup$ Can you further clarify what you mean, i don't quite understand. $\endgroup$ – josh Apr 8 '19 at 12:54
  • $\begingroup$ The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear. $\endgroup$ – Michael Rozenberg Apr 8 '19 at 13:52

Hint: $a,b \in\mathbb Q\implies \dfrac{a\pm b}2\in\mathbb Q$.

Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.

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Sum of rational and irrational numbers is irrational. See here.

If $x,y$ are irrational and $x+y$ is rational, then: $$x-y=(x+y)-2y \ \text{is irrational.}$$

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