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Excuse me for a silly question like this. I am 60 years old retired engineer and want to learn some basic math I di did not learn earlier.

I know an example where $(x + y\pi)$ can be an integer, where $y$ is rational , * means multiplication But how to prove the general case if $x$ is always rational , always non-rational or both depending upon the case.

Thanks

thibo
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  • You have to exclude $Y=0$. With that exclusion, then if your equation held we could evaluate $\pi$ as $\pi = \frac {Z-X}Y$ which would contradict the fact that $\pi$ is not rational. – lulu Feb 22 '21 at 15:51
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    Note: your example had better have $Y=0$ or it can not be correct, for the same reason. – lulu Feb 22 '21 at 15:52
  • Multyplying a non zero rational number by a irrational number is irrrational. The additon of a rational number with an irrational is rational. – ben huni Feb 22 '21 at 17:12
  • Thank you very much dear friends. I would like to evaluate each of the statements. Slow as I am due to my age, it may take a little time though. Thanks for encouraging me. – Shaldar2021 Feb 23 '21 at 05:19
  • Hi lulu, Thanks for your easy proof. I understood it. – Shaldar2021 Feb 23 '21 at 15:29
  • Hello Ben, Thanks. I am little confused with this statement below - is it a typo(italics below)? "The additon of a rational number with an irrational is rational." Addition rules,my guess should be :
    1. rational + irrational =irrational
    2. irrational + irrational = rational or irrational

    My original question ( already proof has been provided by lulu and M.Nestor )points to this case irrational + irrational = rational ( under some constraints, may be )

    – Shaldar2021 Feb 23 '21 at 15:45

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Suppose for a contradiction $x$ is rational. Then $x=a/b$ where $a,b$ are coprime integers and $b\neq0$. Also assume $y=c/d$ for coprime integers $c$ and $d$, and furthermore assume $c$ is not zero (so $y$ is not zero). Then, $$ x = z - \pi y $$ if and only if $$ \frac{a}{b} = z - \pi \frac{c}{d} $$ Multiply by $bd$: $$ ad = zbd - \pi cb $$ Then, $$ \pi cb = zbd - ad $$ Divide by $cb$ (this is possible since both are nonzero): $$ \pi = \frac{zbd-ad}{cb} $$ This implies that $\pi$ is rational which is a contradiction.

  • Great, I will understand the equations better than words. I will revert back soon. Thanks a lot for showing me the way. – Shaldar2021 Feb 23 '21 at 05:21
  • Thank you very much, Nestor. It's beautiful, I never learnt this method of proof by Contradiction. I think I shall utilize similar approach to prove few of my doubts regarding multiplications of irrationals with irrational and rational. I may come back to you in case of doubt. Have a nice day. – Shaldar2021 Feb 23 '21 at 15:50