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Problem: Prove if $ax = a$ for some number $a \neq 0$, then $x=1$.

Here is my proof: If $ax=a$, then $a^{-1}ax=a^{-1}a.$ By the existence of multiplicative inverses, $x=1$. Q.E.D

Now, I have a feeling that this doesn't show that $x$ HAS to be $1$. Is my proof complete?

Levi
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Athelstan Carlton
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3 Answers3

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You should be careful and state why $a^{-1}$ exists before introducing it into the proof. Carefully write out what $a^{-1}a$ means and equals, and deduce that $x = 1$ from the laws of arithmetic that Spivak gives you.

Terra
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If you are working in a division ring (such as a field), what you've done here is sufficient. It should be noted that the statement of the existence of multiplicative inverses should precede the introduction of $a^{-1}$.

Rushabh Mehta
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I'll assume you're working with the real numbers. The reals form a field and so when you remove the zero, the multiplicative inverse exists for each element. What you have done is valid.