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If $a\cdot a = a$ then $a=0$ or $a=1$?

Where $a \in \mathbb{R}.$

I try this way:

Suppose that a is different to $0$ it implies the existence of multiplicative inverse:

$$a^{-1}(a\cdot a)=a^{-1}\cdot a$$

$$(a^{-1}a)= 1~~~~~\text{(by associativity)}$$

$$1\cdot a=1~~~~~\text{(multiplicative inverse)}$$

Therefore $a=1$.

Is this even possible to negate one of my conclusion to get to the other one and then do the same to prove the other conclusion or this is terribly wrong? I appreciate every help I'm losing myself.

John W. Smith
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Adrián Juárez Carrasquedo
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  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 18 '18 at 20:34
  • oh yeahnext time i will do better i'm new here – Adrián Juárez Carrasquedo Sep 18 '18 at 20:36
  • You can [edit] the question, you know? Try MathJax too :) – Shaun Sep 18 '18 at 20:37
  • yeah it is in the real numbers field I forget about that – Adrián Juárez Carrasquedo Sep 18 '18 at 20:57
  • Your argument is fine, but in the presentation you are missing one (trivial) part. You should start by showing that $0\cdot 0 = 0$ proving that $0$ satisfy it. Having done that you can assume $a\not =0$ for which your argument shows that $a=1$. – Winther Sep 18 '18 at 21:16

2 Answers2

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You claim that if $a$ is different from $0$, then it has a multiplicative inverse. This is only true if you're working over a field such as the real numbers. In general, this doesn't have to be true.

A concrete counterexample to your claim is ${\mathbb Z}[X]/(X^2-X)$ and $a$ equal to (the residue class of) $X$. By construction, $a^2 = a$, but $a$ is neither $0$ nor $1$.

Over a field, your reasoning is correct. I'd shorten it to simply: $a^2 = a$ $\Rightarrow$ $a(a-1) = 0$ $\Rightarrow$ $a = 0$ or $a-1 = 0$.

Magdiragdag
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    Please address this: Is this even possible to negate one of my conclusion to get to the other one and then do the same to prove the other conclusion or this is terribly wrong? I think that is the main question, pedagogically. – Shaun Sep 18 '18 at 20:57
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    @Shaun You may be right. I can't look in the OPs mind, though, so I'll wait for clarification. – Magdiragdag Sep 18 '18 at 21:00
  • The field im working on is in the real numbers where a belongs – Adrián Juárez Carrasquedo Sep 18 '18 at 21:04
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A priori you cannot assume there exists a multiplicative inverse.

Instead, using associativity (I hope you have that, at least) rewrite $a^2=a$ as $a\cdot(a-1)=0$. Now it all depends on the fact if your ring allows zero divisors or not. If it does not, then either $a=0$ or $a=1$, otherwise not.

b00n heT
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