We just started number theory and I am little confused on how to prove the statement below:
For $c, d ∈ Z$, prove that if $c \mid d$ and $d\mid c$ then $c = d$ or $c = −d$
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1Please first share what you thought about trying to do. – Daniel W. Farlow Mar 21 '15 at 22:53
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2Hint: $c\mid d\implies |c|\le |d|$. – user26486 Mar 21 '15 at 22:53
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My hint was for $d\neq 0$. If $d=0$, then $d\mid c\implies c=0$, and clearly $0=\pm 0$. – user26486 May 10 '15 at 11:49
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http://math.stackexchange.com/questions/940718/prove-or-disprove-if-a-divides-b-and-b-divides-a-does-a-b – user26486 May 21 '15 at 18:38
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If $c\mid d$ there is some integer $q$ so that $cq=d$. Since $d\mid c$ there is an integer $r$ so that $c = dr$. You then have $d = cq = drq$. Because $d \not= 0$, you have $rq = 1$ for integers $r$ and $q$. Hence $r$ and $q$ are both $1$ or $-1$.
Thomas Andrews
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ncmathsadist
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Assume that $c,d\ne0$. We have $c|d\iff \exists k\in\Bbb Z\mid d=ck$ and similarly $d|c\iff c=k'd$ then by multiplying the two equalities we get $kk'=1\iff k=\pm1$. Conclude.
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Clear if $\,c\,$ or $\,d=0.\,$ Else $\ \dfrac{c}d,\,\dfrac{d}c\ $ are integers with product $1$ so they are $\pm1$
Alternatively $\,\ c\mid d\mid c\,\Rightarrow\ |c| \le |d| \le |c|\,\ $ so $\,\ |c| = |d|$
Alternatively $\,\ |c| \overset{\large c\,\mid\, d_{\phantom{1_{1_1}}}\!\!\!\!\!\!}=\gcd(c,d)\overset{\large d\,\mid\, c_{\phantom{1_{1_1}}}\!\!\!\!\!\!}=|d|$
Bill Dubuque
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