An easy number theory problem

Number Theory Level pending

Positive integers $a$ , $b$ , $c$ , and $d$ are such that $\gcd (a,b)=1$ and $a<b$ , and $\gcd(c,d)=1$ and $c<d$ . Then $\dfrac{a}{b}+\dfrac{c}{d}=1$ is true only if $b=d$ .

False True

1 solution

Kelvin Hong
Jul 1, 2017

Consider a fraction $\dfrac{a}{b}$ ,

$1-\dfrac{a}{b}=\dfrac{b-a}{b}$

$\gcd(b,b-a)=\gcd(b,b-b-a)=\gcd(b,-a)=\gcd(b,a)=1$

So the fraction $\dfrac{b-a}{b}$ cannot divided down simpler and since $\gcd(c,d)=1$ , there must be $b=d$ .

Someone please tell me if there is a theory talk about this statement, thanks.

An easy number theory problem

1 solution

0 pending reports