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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 ...


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.


Note that there are saveral principles in place here. (i) The principle of mathematical induction : $[P(0) \land \forall k (P(k) \rightarrow P(k+1))] \rightarrow \forall n P(n)$. You want to prove a form of (ii) The principle of strong induction : $\forall k[\forall m(m < k \rightarrow P(m)) \rightarrow P(k)] \rightarrow \forall nP(n)$. ...


If $(x|y)\lor(x|z)$ then $(y=kx)\lor(z=k'x)$ so $(yz=(kz)x)\lor(yz=(k'y)x)$ so in the both cases $x|yz$. The contrapositive is the desired result.


If $x\mid y$, then $y = kx$ for some integer $k$. If $x \mid z$, then $z = jx$ for some integer $j$. $$\implies yz = (kz)x \lor yz=(jy)x \implies x\mid yz$$ by the definition of divisibility.

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