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Let's define $\mathbb{N}^*$ first: $\mathbb{N}^* = \mathbb{N} - \{0\}$

Prove if $a$ & $b$ & $d$ $\in \mathbb{N}^*$ and $gcd(a,b)=d$ and $a=d\cdot k$ and $b=d\cdot n$ then $gcd(k,n)=1$.

In order to resolve it, I've done the following:

We have $a=d\cdot k$ and $b=d\cdot n$ so: $$\frac{a}{k}=\frac{b}{n}$$

And we can easily show that if $\zeta \in \mathbb{N}^*$ then: $gcd(\zeta,\zeta)=\zeta$

Thus: $$gcd(a\cdot n, b \cdot k)=d$$

But can I directly do: $$\frac{gcd(a\cdot n, b \cdot k)}{gcd(a,b)=d}=\frac{d}{d} = 1$$ and the $a$, $b$ cancel out so I get the final result: $gcd(k,n)=1$ and why?

Thank you in advance.

Bill Dubuque
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  • Your question is equivalent to the original problem. On the one hand, if you know your question to be true, then you can certainly use it. On the other hand, if you don't know that to be true, then you are assuming the answer to the original problem (which is not okay). – davidlowryduda Oct 19 '13 at 23:13
  • You have $\frac{a}{k} = d = \frac{b}{n}$, so $a\cdot n = b \cdot k = d\cdot k\cdot n$. – Daniel Fischer Oct 19 '13 at 23:13
  • @DanielFischer The last expression has a fault in the end. It's $a\cdot n = b \cdot k = d$ not $a\cdot n = b \cdot k = d\cdot k\cdot n$. –  Oct 19 '13 at 23:15
  • @mixedmath Well I know it's true but I have to prove that it is true without using induction. –  Oct 19 '13 at 23:16
  • @Adobe $a\cdot n = \frac{a}{k}\cdot k\cdot n$, and by assumption, $a = d\cdot k$, so $\frac{a}{k} = d$. – Daniel Fischer Oct 19 '13 at 23:18
  • @DanielFischer Ah! Sorry... (tired) But how can I use that result to prove the content of the question? –  Oct 19 '13 at 23:25
  • Let $g = \gcd(k,n)$. What can you say about $d\cdot g$? – Daniel Fischer Oct 19 '13 at 23:27
  • Hmm.. I have no idea... I'm locked at that point. –  Oct 19 '13 at 23:36

1 Answers1

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Assume that $gcd(a,b)=d$. This leads to conclusion that $a=k \cdot d$ and $b = n \cdot d$

Now let $gcd(k,n) = m$, then we can write: $k = s \cdot m$ and $n = t \cdot m$. Substitute back we have: $a = s \cdot m \cdot d$ and $b = t \cdot m \cdot d$. Both $a$ and $b$ have $md$ asa factor, but we know that $gcd(a,b) = d$, so that implies $m=1$

Q.E.D.

Also another way is to use the following property of the function Greatest Common Divisor:

$$gcd(ma,mb) = m \cdot gcd(a,b) \forall m \in \mathbb{Z}; m \neq 0$$

Let $a = kd$ and $b=nd$. Then from assumtion we have:

$$gcd(a,b) = d$$ $$gcd(kd,nd) = d$$ $$gcd(k,n) \cdot d = d$$ $$gcd(k,n) = 1$$

Q.E.D.

To make it clear the Greatest Common Divisor is as I would like to call it "semi-multiplicative" function, i.e. it's multiplicative, but only under specific condition. For example, this will hold:

$$gcd(ma,b) = gcd(m,b) \cdot gcd(a,b)$$

if $gcd(m,a) = 1$

So your calculation is wrong because it true that:

$$gcd(\frac ak, \frac bn) = d$$

But you multiply both sides with $nk$, that will change the value of The GCD, unless $(n,k) = (a,n) = (b,k) = 1$, but we can't make that restriction.

Also that division will never lead to an answe that $gcd(n,k)=1$. Because we can make the following factorization if $(a,n) = (b,k) = 1$

$$gcd(a \cdot n, b \cdot k) = (a,b) \cdot (a,k) \cdot (b,n) \cdot (n,k)$$

But obviouly we can't make that restriction. For counterxample take $a=12$ and $b=18$.This will provide $d=6$, $k=2$, $n=3$.

So in your example you'll have:

$$(a\cdot n, b \cdot k) = (36,36) = 36 \neq d = 6$$

Stefan4024
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