The question I am working on is:

Prove that if x is irrational, then 1/x is irrational.

My proof differs from the one given in the answer key; but I still feel that mine is valid. Could someone possibly look over my proof to see if it is correct?

Proof by contraposition: If $1/x$ is rational, then x is a rational number.

Assuming that $x \ne 0$, then $1/x$ is by definition a rational number; taking the reciprocal of this, x is will be some number, other than zero, that can be written as $x/1$, which is a rational number by definition.

Since we have proven the contrapositive to be true, then the original statement must be true.

EDIT: I found this solution on the internet.

Proof: We prove the contrapositive: If 1=x is rational, then x is rational. So suppose 1=x is rational. Then there exist integers p; q, with q = 0, such that 1 6 =x = p=q. Then x = q=p is clearly rational, unless p = 0. However, the case that p = 0 can't occur, because if p = 0, then 1=x = p=q = 0. But 1=x is never zero.

My question is, what is the point in mentioning the case that $p=0$. Isn't it safe to assume that, once you reach the point when you take the reciprocal, $p$ can't equal zero??

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    $\begingroup$ Why is $x/1$ rational by definition? $\endgroup$ – David Mitra Feb 14 '13 at 14:05
  • $\begingroup$ @DavidMitra Because it is written as a fraction. $\endgroup$ – Mack Feb 14 '13 at 14:07
  • $\begingroup$ With $x$ an integer? With $x$ rational? It seems you're assuming the result here. $\endgroup$ – David Mitra Feb 14 '13 at 14:08
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    $\begingroup$ Any number $y$ can be written as $y/1$. Is $y\ne 0$ always rational? The last clause in your proof needs to say a bit more: you want to say that $x$ can be written as $p/q$ where $p$ and $q$ are integers (as in Dominic's answer). $\endgroup$ – David Mitra Feb 14 '13 at 14:17
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    $\begingroup$ @EliMackenzie $\pi /2$ is also written as a fraction but is irrational. $\endgroup$ – alancalvitti Feb 14 '13 at 15:21

Yeah that should do it, even though I would prefer something like for $x\neq 0$ $$ \frac{1}{x}=\frac{p}{q} \iff x= \frac{q}{p} $$ Edit: Should first mention that $x$ ist rational

  • $\begingroup$ That proof is actually what the answer key provides. $\endgroup$ – Mack Feb 14 '13 at 14:06
  • $\begingroup$ Your contraposition isn't a real contraposition, because of $$\frac{1}{\frac{1}{x}}=x$$ $\endgroup$ – Dominic Michaelis Feb 14 '13 at 14:09
  • $\begingroup$ Because of what? $\endgroup$ – Mack Feb 14 '13 at 14:09
  • $\begingroup$ With taking $z=\frac{1}{x}$ you see that if $z$ irrational implies that $\frac{1}{z}$ is irrational, than $\frac{1}{x}$ irrational implies that $x$ is irrational. So the proof by contraposition is a constructive one $\endgroup$ – Dominic Michaelis Feb 14 '13 at 14:16

The big gap in your argument is that you said $x/1$ is a rational number "by defnition". The definition says it's an integer over another integer. You can't conclude $x$ is an integer, so $x/1$ has not been shown to be an integer over and integer.

If $1/x$ is rational, then $1/x = m/n$ where $m$ and $n$ are integers. It follows that $x=n/m$. Since $m$ and $n$ are integers, this is rational.


For Contradiction Let us assume that $ \frac {1}{x}$ is rational i.e $ \frac {1}{x}$=$ \frac {p}{q}$ ($q \neq 0$) and more over $p \neq 0$ since $x \neq 0$ this implies that $x= \frac{q}{p}$ ($p\neq 0$) is rational. this is the contradiction to our assumption that $ \frac{1}{x}$ is rational.


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