0

I am solving this exercise problem from Hernstien it goes like:

If $a= p_1^{\alpha_ 1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and $b =p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$

Prove that: $(a,b)=p_1^{\delta_1}p_2^{\delta_2}\cdots p_k^{\delta_k})$, where $\delta_i = \min(\alpha_i,\beta_i)$ for all $i$.

My attempt:

Proof: Since, $p_i>0$ for all $i$, ${p_i}^{\delta_i}$ should be as large as possible.

Since, $a= p_1^{\alpha_ 1}p_2^{\alpha_2}\cdots p_k^{\alpha_k},$ $p_1^{\alpha_1}\mid a$ and similarly, $p_1^{\beta_1}\mid b,$ which implies $p_1^{\min({\alpha_1,\beta_1})}\mid a$ and $p_1^{\min({\alpha_1,\beta_1})}\mid b$ $\implies$ $p_1^{\delta_1}$ divides both $a$ and $b.$
Therefore, $p_i^{\delta_i}$ divides both $a$ and $b$ for all $i.$
Thus, $(a,b)=p_1^{\delta_1}p_2^{\delta_2}\cdots p_k^{\delta_k}.$

Is my proof correct? Any improvements are welcome.

Bill Dubuque
  • 272,048
Priyanshu
  • 1
  • You've only shown that $p_1^{\delta_1}\cdots p_k^{\delta_k}$ is a factor of $\gcd(a,b).$ – Thomas Andrews Sep 01 '22 at 17:07
  • You really don't need all those parentheses. – Thomas Andrews Sep 01 '22 at 17:08
  • 1
    As a rule, don't use $\implies$ and $\forall$ as word substitutes. They belong in formal statements, and on blackboards to save space, but these are not formal statements and not on a blackboard. Just use the words "implies" and "for all." – Thomas Andrews Sep 01 '22 at 17:11
  • @Thomas Andrews I tried it again but I'm still unable to prove it – Priyanshu Sep 01 '22 at 17:23
  • This is one of those proofs which is easier to "see" than prove exactly. But one approach is to first show that if $\gcd(b_1,b_2)=1$ then $\gcd(a,b_1b_2)=\gcd(a,b_1)\gcd(a,b_2).$ Then prove your theorem inductively on $k.$ – Thomas Andrews Sep 01 '22 at 17:34
  • Neither of these duplicates actually proves this theorem. The first just uses the theorem, without proof. I'm sure it is a duplicate, but getting the right duplicate would be useful for future searches @billdubuque – Thomas Andrews Sep 01 '22 at 17:40
  • @Thomas No, there are proofs in (some) answers there, and of course there are many other dupe targets (all essentially the same as the proofs linked). – Bill Dubuque Sep 01 '22 at 17:45
  • You can complete your proof with the theorem: If $d\mid a$ and $d\mid b$ then $d=\gcd(a,b)$ if and only if $\frac{a}d$ and $\frac bd$ have no common factor other than $1.$ In your case, $\frac{a}{d}=p_1^{\alpha_1-\min(\alpha_1,\beta_1)}\cdots p_k^{\alpha_k-\min(\alpha_k,\beta_k)},$ and similarly for $b.$ Then show that $\alpha-\min(\alpha,\beta)\neq 0$ implies $\beta-\min(\alpha,\beta)=0.$ So $\frac ad$ and $\frac bd$ have no common prime factor, and thus no common factor other than $1.$ – Thomas Andrews Sep 01 '22 at 17:55

0 Answers0