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Let $k$ be a positive integer and $x=\sqrt{k}$. Suppose $x$ is rational and $x=( \frac{m}{n} )$ such that $m\in\mathbb{Z}$ and $n$ is the least positive integer such that $nx$ is an integer. Define $n'=n(x-[x])$ where $[x]$ is the integer part of $x$.

$(a)$ Show that $0\leq n' <n$ and $n'x$ is an integer.

$(b)$ Show that $n'=0$

$(c)$ From $(a)$ and $(b)$, conclude that $\sqrt{k}$ is either a positive integer or irrational.

In this problem, I could establish the inequality using properties of fractional part of x. However, since the question does not mention any such function, I don't think it should be used. Can someone help me solve this.

Bill Dubuque
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Rudransh Goel
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    you can use division algorithm to write $m = qn + r$ – YOu will not know Jan 05 '22 at 10:32
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    $(x-[x])$ is the fractional part of $x$ and is $\in [0,1)$ – Henry Jan 05 '22 at 10:32
  • Since $n$ is positive, $n'=0$ immediately implies that $x$ must be equal to its integer part which is only possible if $x$ is an integer. – Peter Jan 05 '22 at 10:48
  • The definition of $n$ ensures that $m$ and $n$ are coprime – Peter Jan 05 '22 at 10:53
  • See John Conway's proof in this answer for this method of proof (done slightly more slickly), and follows the links there for much more (see esp. Theorem 2 in the 2nd dupe, which is precisely in the form of your exercise, and the discussion there of this form of proof). – Bill Dubuque Jan 05 '22 at 11:00
  • See also here and here. – Bill Dubuque Jan 05 '22 at 11:07
  • If you read the linked proofs, you will see how the proof can be viewed as a descent on the denominators for $x$, i.e. the arguments show that if $n>1$ is a denominator for $x$ (i.e. $nx\in\Bbb Z),$ then $x$ can also be written with a smaller denominator $n'.,$ This yields a contradiction if the least denom $n$ is $> 1,$ (or, equivalently, it yields an infinite descent on denominators $, \ldots n'' < n' < n,,$ contra $\Bbb N$ is well-ordered). Thus the least denom must be $,n=1,,$ i.e. $,x\in\Bbb Z\ \ $ – Bill Dubuque Jan 05 '22 at 11:18
  • If anything remains unclear please post further comments (either here or on the linked answers). – Bill Dubuque Jan 05 '22 at 11:21

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