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Prove: If $a$ is an integer and $|a| > 1$, then $1/a$ is not an integer.

Hi, I need help proving this either by contradiction or contrapositive. I'm not sure where to begin

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  • $\begingroup$ It's not true. Suppose $a \in \mathbb Z = 3 >1$. $\;\frac 1a = \frac 13 \notin \mathbb Z$ $\endgroup$
    – amWhy
    Jul 9 '14 at 16:30
  • $\begingroup$ I'm not sure if I'm not understanding it correctly: $2\in \mathbb{Z}$, $\vert 2 \vert >1$, but $\frac{1}{2} \notin \mathbb{Z}$. $\endgroup$ Jul 9 '14 at 16:30
  • $\begingroup$ The statement is blatantly false if $Z$ is meant to be the integers. Do you perhaps mean it to be something else? $\endgroup$ Jul 9 '14 at 16:30
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    $\begingroup$ May be the question is $\dfrac{1}{a} \notin \mathbb{Z}$. $\endgroup$
    – hrkrshnn
    Jul 9 '14 at 16:31
  • $\begingroup$ @boywholived sorry, you are correct. how would i do that with a formal proof? $\endgroup$
    – Sam
    Jul 9 '14 at 16:36
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Suppose $\frac{1}{a} = b$ an integer. I multiply both sides of this equation by a, giving $ab = 1.$ This implies $a = b = \pm 1$ since the only integers who multiply out to $1$ are $\pm 1$ (since they are the units in the ring of integers) . But $|a|>1$. This is a contradiction.

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  • $\begingroup$ im sorry, for some reason i can't seem to understand this. are you trying to say that 1/a is an integer? $\endgroup$
    – Sam
    Jul 9 '14 at 16:40
  • $\begingroup$ Yes, this a proof by contradiction. I suppose that 1/a is an integer b and, using this, I reach the contradiction that |a| = 1. But, this can't be the case since you stipulated |a| >1, strictly. Hence, the assumption that 1/a is an integer is false. $\endgroup$
    – MRicci
    Jul 9 '14 at 16:42
  • $\begingroup$ okay, and then how did you get this part? ab=1⟹a=b=±1 (why is a=b) $\endgroup$
    – Sam
    Jul 9 '14 at 16:43
  • $\begingroup$ could we also prove it using this logic? Assume a ∈ ℤ, then a > 1 or a < -1. But if a =3, then 1/3 ∉ ℤ. By definition of integer, this is a contradiction. $\endgroup$
    – Sam
    Jul 9 '14 at 16:45
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    $\begingroup$ @Sam what you just wrote is an example of how this is not true, not a full on proof. For example I can come up with an example of integers $a,b$ where $a\neq b$ but $ab = 0$, i.e. $a = 5, b = 0$ but this doesn't prove that for every pair of integers $ab = 0$. $\endgroup$
    – DanZimm
    Jul 9 '14 at 17:08
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If $|a|>1$ then $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So, $\frac{1}{|a|}$ is a number between $0$ and $1$. Since there is no integer between $0$ and $1$, we conclude that $\frac{1}{|a|}\notin\mathbb Z$.

EDIT: Proof by contradiction, as you want:

Suppose $\frac{1}{|a|}$ is an integer. Since $|a|>1$, we conclude that $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So, $\frac{1}{|a|}$ is an integer between $0$ and $1$. But it's a contradiction, because there is no integer between $0$ and $1$. It follows that $\frac{1}{|a|}\notin\mathbb Z$ and thus $\frac{1}{a}\notin\mathbb Z$ (because $\frac{1}{a}$ is an integer if and only if $\frac{1}{|a|}$ is an integer)

EDIT 2: (to make my answer valid as pointed out by @TomOldfield)

There is no integer between 0 and 1.
Proof: Suppose that there exists a number $p\in\mathbb Z\cap (0,1)$. Let $(x_n)$ be a sequence defined by $x_n=p^n$. Notice that $x_n\in\mathbb Z\cap (0,1)$ for all $n\in \mathbb N$. Since $(x_n)$ is strictly decreasing, it follows that $\mathbb Z\cap (0,1)$ does not have a least element. But it contradicts the well-ordering principle. So, $\mathbb Z\cap (0,1)=\varnothing$.

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  • $\begingroup$ Saying that there are no integers between $0$ and $1$ is essentially just a restatement of the question. Unless you prove that there are no integers between $0$ and $1$, I don't think this is a valid answer. $\endgroup$ Jul 11 '14 at 16:09
  • $\begingroup$ @TomOldfield For that fact (that is just a restatement of the problem), there are a lot of of references on the internet (for example, on this forum). However, to make a valid answer for people who think like you, I've edited my post. $\endgroup$
    – Pedro
    Jul 12 '14 at 21:35
  • $\begingroup$ Beautiful, +1! Perhaps to clarify, I think that it is clear that your first part was a restatement of the problem, and that was exactly why I didn't like it, it was trivially a restatement so that it seemed like no work had been done. But the edit uses a fundamental property of $\mathbb{N}$ to actually prove the restatement, which then proves the desired fact. Even better, it's totally different to how I would have approached it, so I got to learn something too! $\endgroup$ Jul 12 '14 at 23:58

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