2

$\sqrt {3} \in Q$. Then, $\sqrt{3} = \frac ab$ with the lowest term for $a,b \in Z$.

Then, $3b^2=a^2$, which implies that $a^2$ is divisible by 3.

That is, $a$ is also divisible by 3 (by fundamental theorem of arithmetic).

I don't understand here $a^2$ divisible by 3 implies $a$ divisible by 3.

Could you explain it?

shk910
  • 3,599

2 Answers2

2

Working with integers, recall that $cd$ divisible by a prime $k$ implies $c$ is divisible by $k$ or $d$ is divisible by $k$. $a^2$ divisible by 3 implies $a \cdot a$ divisible by 3. So, $a$ is divisible by $3$ or $a$ is divisible by $3$ (This statement occurs by taking the left $a$ and the right $a$; if you don't understand this, in the example mentioned let $c = a$ and $d=a$).

Compact
  • 440
  • 2
    cd divisible by k implies c is divisible by k or d is divisible by k Only if $k$ is a prime.Otherwise take $c=d, k=c^2$ for example. – dxiv Mar 06 '18 at 00:27
  • It's simple. Thanks. – shk910 Mar 06 '18 at 00:29
  • 1
    @SihyunKim Be careful! This answer is wrong. –  Mar 06 '18 at 00:33
  • why is it wrong?? – shk910 Mar 06 '18 at 00:39
  • What am I missing here? Last time I checked, 3 was still prime. – MJD Mar 06 '18 at 00:45
  • @MJD The original version of the answer said that $k | cd \implies k | c$ or $k | d$ unconditionally; it made no use of the fact that $3$ is prime. –  Mar 06 '18 at 22:35
  • But since the theorem was applicable in this instance, the answer was correct. Even if the answer omitted to mention one of the conditions, the condition was still satisfied. In my opinion your claim that "this answer is wrong" was wrong. – MJD Mar 06 '18 at 23:41
1

Consider the prime factorization of $a$.

Suppose on the contrary that $3$ is not a factor of $a$. Then squaring $a$ will also not make $3$ appears in the prime factorization of $a^2$. Hence $3$ will not be a factor of $a^2$.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149