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There is known the non-constructive proof for proving the above, but the inverse is not possible to be proved in the same (non-constructive approach).

The inverse would be : $(a,b)>1, a\mid c, b\mid c \nrightarrow ab\mid c$
Employing the indirect, or the contradiction approach, assume that $ab\mid c$.
Also, $\exists k,l \in \mathbb {Z+}, c=ka, c= lb.$
Next, let $\exists n \in \mathbb {Z+}, (a,b)=n \implies \exists a',b' \in \mathbb {Z+}, a= na', b=nb'$.
$c = ka'n = lb'n$.
But, $ab = a'b'n^2$; & nothing can be stated using this sort of argument about the relation between $a'b'n^2, ka'n$, or between $a'b'n^2, lb'n.$
If say $a'b'n^2 \nmid c$, then it is improper as nothing is known about the relation between the two terms as stated above.

So, what can be a proper way to describe the failure of this way of proving the inverse.

If need can give some example to support the obvious fact about 'any' sort of relationship that can occur between $ab=a'b'n^2$ & $c=ka'n=lb'n.$

jitender
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  • This very question was asked and perfectly answered earlier today. – Arnaud Mortier Feb 13 '18 at 22:16
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    $a=b=2$ and $c=4$, one has $(a,b)=2,a\mid c,b\mid c$ and $ab\mid c$. On the other hand $a=b=c=2$ one has $(a,b)=2,a\mid c,b\mid c$ but $ab\nmid c$. If you are trying to prove that $(a,b)>1,a\mid c,b\mid c$ doesn't imply that $ab\mid c$ all you need is an example. If you are trying to prove that $(a,b)>1,a\mid c, b\mid c$ implies that $ab\nmid c$ that is false by my earlier counterexample. – JMoravitz Feb 13 '18 at 22:16
  • This question does not make sense. Obviously it is not true that $\gcd(a,b)>1, , a,|,c, , b,|,c\implies ab,|,c$, So what is the point? – lulu Feb 13 '18 at 22:18
  • @JMoravitz So, there cannot be a non-constructive proof, and a constructive proof is needed to show the both cases as in your example. – jitender Feb 13 '18 at 22:18
  • @lulu But, then need a constructive example to first know if a non-constructive proof is possible or not. I have not read this in books on logic, and request you to please guide on this aspect of pre-checking by an example. – jitender Feb 13 '18 at 22:21
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    First of all, this is not the inverse statement. Secondly, I don't get the emphasis on constructive vs. non-constructive in this context. Can you give me a non-constructive proof that $4$ isn't prime? – lulu Feb 13 '18 at 22:23
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    The inverse is not true, since it would require the converse to be true, and the converse is not true. – Thomas Andrews Feb 13 '18 at 22:23
  • @ThomasAndrews I have some doubt about correctly using the word 'inverse', that was chosen for inverting only one condition, among three available. The input conditions are : $a\mid c, b\mid c., (a,b)=1$. I took only one condition; & it might be erroneous.But, I felt that taking the divisibility condition on the l.h.s can be taken as constant. – jitender Feb 13 '18 at 22:23
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    The inverse of $P\implies Q$ is $\lnot P\implies\lnot Q$, which is equivalent to $Q\implies P$, which is not true here because $ab\mid c$ does not imply that $\gcd(a,b)=1,$ which is part of $P.$ – Thomas Andrews Feb 13 '18 at 22:26
  • @ThomasAndrews I got it, the converse itself is missing on the necessity of the divisibility of $c$ by $ab$ to ensure co-primeness of $a,b$. – jitender Feb 13 '18 at 22:28
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    Now, if you wrote it as $$\forall a,b, (\gcd(a,b)=1\implies \forall c((a\mid c\land b\mid c)\implies ab \mid c))$$ then the inverse is true when you take the expression $P$ to be $\gcd(a,b)=1$ and the condition $Q$ to be $ \forall c((a\mid c\land b\mid c)\implies ab \mid c)).$ – Thomas Andrews Feb 13 '18 at 22:29
  • @ThomasAndrews I feel that your modification is the 'only' way to make the inverse possible, as now by taking the co-primeness as fixed, the converse is also true. That would definitely serve as a check to select the 'proper' fixed conditions. – jitender Feb 13 '18 at 22:37
  • I wish that somebody posts an answer or upvote it, as for me it is a very important question, and must be prevented from automatic deletion due to down-voting. – jitender Feb 13 '18 at 22:41

1 Answers1

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If you wrote your statement as:

$$\gcd(a,b)=1\implies\left(\forall c\left(a\mid c\land b\mid c\right)\implies ab\mid c\right)$$

Now, the inverse statement is a constructive consequence of:

$$\gcd(a,b)\neq 1\implies\exists c\left(a\mid c\land b\mid c\land ab\not\mid c\right)$$

That's because $\lnot \forall c(P(a,b,c))$ is a consequence of $\exists c(\lnot P(a,b,c))$ . and $\lnot(X\implies Y)$ is a consequence of $X\land \lnot Y.$

So, given $\gcd(a,b)\neq 1$, you can use $c=\frac{ab}{\gcd(a,b)}$ and deduce that $ab\not\mid c$ but $a\mid c$ and $b\mid c.$

So this constructively finds $c$, but the real question is whether the statement that $\lnot(X\implies Y)$ following from $X\land \lnot Y$ is constructive.

Thomas Andrews
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  • Thanks a lot, but am confused about the last line, that states that can find value of $c$ constructively, and asks if the statement $\lnot(X\implies Y)$ is constructive or not. I hope it refers to finding $ab \nmid c$ constructively, or non-constructively; and hope that this again needs only two constructive cases to prove. I feel that this sort of constructive cases serve as an axiom, that must be needed for a non-constructive proof. I may not be correct about the word 'axiom' in this context, but could not find a better one. – jitender Feb 13 '18 at 22:58
  • I hope my earlier comment regarding the last line of your answer is fine, except for usage of the word 'axiom'. – jitender Feb 13 '18 at 23:20
  • Sorry, but I am getting confused and skeptical when consider your statement : "you can use $c= \frac {ab}{gcd(a,b)}$", as there is not a clear cut relation like that, as $c = ka'n=lb'n, ab = a'b'n^2$, with $n= (a,b)$. – jitender Feb 15 '18 at 13:59
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    No idea what you mean. This $c$ is divisible by $a$ and it is divisible by $b$, but it is not divisible by $ab$ since it is smaller than $a,b.$ – Thomas Andrews Feb 15 '18 at 14:29
  • Enough for what? – Thomas Andrews Feb 15 '18 at 14:33
  • But, then the ratio expressed for $c$ must be having a factor multiplied to it too, as division by $\gcd(a,b)$ does not seem enough for expressing $c$. I hope that $c =$ (some integer value, denoted by let, $z$) * $\frac{ab}{gcd(a,b)}$ would better express the value of $c$ . – jitender Feb 15 '18 at 14:37
  • I request some help for my post, as it was driven by your answer and comments, and is stuck at the last step. It is at: https://math.stackexchange.com/questions/2651730/prove-for-a-prime-p-forall-k-in-bbb-z-a-equiv-b-pmod-p-nrightarrow-a-e – jitender Feb 15 '18 at 14:44
  • Have an EDITED version of the proof stated in the earlier comment, please help. https://math.stackexchange.com/questions/2651730/prove-for-a-prime-p-forall-l-in-mathbb-z-a-equiv-b-pmod-p-nrightarrow – jitender Feb 15 '18 at 16:29