I have a number theory question that has me stumped.
Let $a, b, c \in \Bbb Z$ with $a$ and $b$ both not zero. Prove: $(\frac{a}{(a,b)}, \frac{b}{(a,b)}) = 1$.
Any help would be greatly appreciated!
Asked
Active
Viewed 86 times
2
-
3$(k a, k b) = k (a, b)$ – vonbrand Apr 04 '13 at 00:42
-
What do you know? An answer will depend on what stage you’re at in number theory. – Lubin Apr 04 '13 at 00:43
-
1Please make titles more informative. See what I changed it to, so you can do something similar next time. – Pedro Apr 04 '13 at 00:51
-
1Where is $c$ used? – Thomas Andrews Apr 04 '13 at 01:03
-
It is the speed of light, rounded down... – copper.hat Apr 04 '13 at 01:44
-
Thank you for your help everyone! Mr. Tamaroff, I'll use that in the future! – Neurax Apr 04 '13 at 02:20
-
http://math.stackexchange.com/questions/202397 – Bart Michels Feb 21 '15 at 16:47
4 Answers
6
@vonbrand's comment is the most succinct answer. Here is a more verbose approach:
Let $d =\gcd(a,b)$. Let $l = \gcd(\frac{a}{d}, \frac{b}{d})$. Then $l \mid \frac{a}{d}$, $l \mid \frac{b}{d}$, which implies $ld \mid a$, $ld \mid b$. Hence $ld \le d$, from which we get $l \le 1$. Since $1 \mid a$, $1 \mid b$, we have $1 \le l$. So $l=1$.
copper.hat
- 172,524
1
Bezout's identity will give you the answer immediately: http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
It tells us that there exist integers $x,y$ such that $$xa+yb=(a,b).$$Divided by $(a,b)$ gives $x\frac{a}{(a,b)}+y\frac{b}{(a,b)}=1$, and note that the GCD of $\frac{a}{(a,b)}$ and $\frac{b}{(a,b)}$ must divide the RHS.
Easy
- 4,485
