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I'm learning about axioms of field $\Bbb R$ so I should be able to prove that $1$ is a unique neutral element for multiplication. I'm not entirely sure how to do it so I hope you can help me.

Let's assume there exists $a\in \Bbb R, a\neq 1$ such that $\forall x\in\Bbb R$ we have $ax=xa=x$. How do I now get a contradiction? I can't just simply say now we see $a=1$, can I?

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    $a=a\cdot 1=1$ by the properties of both. – 2017-02-04
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    @AdamHughes: How does that help Lewis find the solution by himself? – 2017-02-04
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    @user21820 it doesn't. It's a solution to a one-step problem that's too short to make me feel I deserve upvotes for posting as a full answer. – 2017-02-04
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    @AdamHughes: Oh okay. When giving hints I try to give the crucial bits and leave out all the rest. But since you've already given the full answer, my answer is no longer of any use. – 2017-02-04
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    @user21820 I agree with this philosophy, it's a good thing, and usually my comments vs answers are just that, I just had the aforementioned lack of entitlement to upvotes feeling and the only way to avoid that is to post as something that doesn't give rep. Cheers! :) – 2017-02-04
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    @AdamHughes: You may be interested in the curious fact that you can prove uniqueness of identity of a group (not just for fields) given only left-identity and left-inverses, that is, $\forall x\ ( 1x=x \land \exists y\ ( yx=1 ) )$, besides associativity. – 2017-02-04
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    @user21820 yes, I remember that problem seeming terribly difficult back during my first course in algebra. Just goes to show how math is sometimes something you get used to over the years rather than outright "learn" in the usual sense. :) – 2017-02-04
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    @AdamHughes: I couldn't solve it when I first encountered it. Hope it presents a challenging follow-up exercise for Lewis. Hint: the solution is short. =) – 2017-02-04

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