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Let me begin by saying that the answer to part of this question has been previously addressed HERE: Prove/Disprove: if $x^2 = a^2$, then $x = a$

However, I am hoping to add something to this inquiry that I think is deserving of a new post all together.

The first part of this question is to determine whether the following statement is true or false for all groups with property of $\ \ x^2=a^2$:

$x^2=a^2 \rightarrow x=a$

Starting with what is assumed:

$x\circ x = a \circ a \ \ \ \ \ \ \ $Then, performing some manipulation...

$ x\circ x \circ x^{-1} = a \circ a \circ x^{-1} $

$x \circ e = a \circ a \circ x^{-1}$

$x = a \circ a \circ x^{-1}$

Now, the only way to show that $x = a$ would be if $a \circ x^{-1} = e$

This would mean that $ a = (x^{-1})^{-1}$ but $(x^{-1})^{-1}$ is just another way of writing $x$.

As such, we need $a = x$ in order to prove $a=x$

Said differently, proving the claim requires the assertion of the claim.

I know a little bit of formal logic to understand that this cannot be..."correct".

Is there a particular name for this type of logical fallacy? And, more importantly, if during proof solving (for any given proof) I encounter this situation, can I immediately conclude that the statement is always false no matter how I choose to tackle it?

Said differently, I know that proofs can be attacked using several different methods (contrapositive, contradiction, induction, etc). If I choose any single proof method and the following phenomenon arises, (i.e. proving the claim requires the assertion of the claim), do I know for certain that there will be NO WAY of ever finding a proof for the claim regardless of if I try a different method?

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Said differently, proving the claim requires the assertion of the claim.

Your reasoning is off here. It isn't accurate to say that proving the claim requires the assertion of the claim. Maybe there is a way to prove it otherwise. (You make a similar mistake in reasoning when you write that "the only way to show that $x=a$ would be...", as Bram28 points out.)

Is there a particular name for this type of logical fallacy?

"Begging the question", or "petitio principii".

If I encounter this situation, can I immediately conclude that the statement is always false no matter how I choose to tackle it?

No. Take any true statement. Now try to prove it, starting by assuming it is true. This is a logical fallacy; you haven't actually done anything. But that doesn't change the fact that the statement is true.

If you could derive a conclusion about the truth of the statement based on this fallacy, then it wouldn't be a fallacy at all, but rather an unbelievably efficient way to know anything. You would effectively be omniscient. In fact, you would be omnipotent as well, since you would be able to decide the truth of anything based on whether you chose to beg the question or not.

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  • $\begingroup$ So, for clarification, the disproof of a claim requires a valid argument that arrives at the "disproof". A petitio principii is, by itself, not a disproof, but rather, an invalid argument. Is that correct? $\endgroup$ – S.Cramer Oct 16 at 22:15
  • $\begingroup$ @S.Cramer Yes, exactly. $\endgroup$ – Théophile Oct 17 at 4:34
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$x = a \circ a \circ x^{-1}$

Now, the only way to show that $x = a$ would be if $a \circ x^{-1} = e$

I question this. Yes, it is true that if $a \circ x^{-1} = e$ then we can show $x = a$ ... but it is not at all clear why showing that $a \circ x^{-1} = e$ is the 'only' way to show $x=a$.

And yes, you're right, assuming that $a \circ x^{-1} = e$ in order to show that $x=a$ would be circular reasoning, or begging the question, which is a fallacy.

But the larger fallacy is to conclude that just because there is some fallacious argument for showing that something is the case, that then that something is suddenly not the case.

For example, here is a really bad argument for the existence of God:

Bananas are yellow Therefore, God exists

OK, bad argument. But that clearly does not mean that God exists is now a false statement.

In your case, you found that with a circular (i.e. bad) argument we can show that $x=a$ if $x^2=a^2$ ... but that does not mean that this statement is now false ... maybe there is a good argument for this.

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  • $\begingroup$ ahhh, great. I had thought about the word choice when I wrote "only". If this is NOT the only way, is there another? I guess showing that both $aa=e$ and $x^{-1}=x$ would be another such way, yes? $\endgroup$ – S.Cramer Oct 16 at 22:11
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    $\begingroup$ @S.Cramer Sure ... if you can show that :) $\endgroup$ – Bram28 Oct 16 at 22:12
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The problem you are discussing is called 'circular reasoning' or 'assuming what you are trying to prove'. If you are interested in a discussion about mathematical reasoning and possible problems with it more generally, I recommend Chapter 1 of Terry Tao's Compactness and Contradiction. On page 11, he explicitly discusses circular arguments, as well as some valid arguments which look circular but actually arent.

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The fallacy is just called a circular argument, but you can definitely not conclude the resulting statement is false.

For example, if we restricting to groups of odd order then the statement is true, even though the circular argument is not valid.

The problem is that just because you can't find a valid argument doesn't mean one doesn't exist.

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  • $\begingroup$ So, for clarification, the disproof of a claim requires a valid argument that arrives at the "disproof". A circular argument is, by itself, not a disproof, but rather, an invalid argument. Is that correct? $\endgroup$ – S.Cramer Oct 16 at 22:15
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    $\begingroup$ Correct, it is just invalid and bears no relevance to whether the statement is true. Typically the easiest way to disprove a statement is to find a counterexample. You could also assume the statement is true and use it to reach a contradiction (this is a proof of the negative of the statement by contradiction) $\endgroup$ – Robert Chamberlain Oct 16 at 22:19

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