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In order to explain what I'm asking, let's consider the following binary operation:

The binary operation $*$ on $\mathbb{R}$ give by $x*y = x+y - 7$ for all $x,y$ $\in \mathbb{R}.$

In here it is pretty clear that the identity element exists and it is $7$, but in order to prove that the binary operation has the identity element $7$, first we have to prove the existence of an identity element than find what it is.

So, how can we prove that the existance of the identity element ?

Note: I actually asked a similar question before, but in that case the binary operation that I gave didn't have an identity element, so, as you can see from the answer, we directly proved with the method of contradiction.Therefore, instead of asking a new question, I'm editing my old question.

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    how is zero the identity element? $x*0 = 3x\ne x.$ – 2017-01-07
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    checked, still confused. But appears others are fielding it. – 2017-01-07
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    I saw my mistake. – 2017-01-07
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    ok (note that it $is$ associative now though) – 2017-01-07

2 Answers 2

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You guessed that the number $7$ acts as identity for the operation $*$. Then you checked that indeed $x*7=7*x=x$ for all $x$. Therewith you have a full proof that an identity element exists, and that $7$ is this special element.

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Remark: the binary operation for the old question was $x*y = 3(x+y)$.

Suppose on the contrary that identity exists and let's call it $e$.

$$0*e = 3(0+e)= 0$$

$$3e=0$$

and hence $$e=0$$

However, $$1*e = 3(1+0)=1$$

and we obtain $$3=1$$ which is a contradiction.

Edit in response to the new question : $\forall x \in Q$, $x + 0 = x$ and $0+x= x$. Hence $0$ is the additive identity.

If you are willing to accept $0$ to be the additive identity for the integer and $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$.

$\frac{a}{b}+\frac{0}{1}=\frac{a(1)+b(0)}{b(1)}=\frac{a}{b}$

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    3(0+e) = 0 ?, I think you are missing something – 2017-01-07
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    what is the definition of identity element? Do you agree that $0*e=0$? do you agree that $0*e=3(0+e)$? – 2017-01-07
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    $x*e = x$ and $e*x = x$, but in the part $3(0+e)$, it is a normal addition. – 2017-01-07
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    Ok, I got it, we assumed that e is exists. – 2017-01-07
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    I edited the definition of the question because, as I pointed out in the question, I'm trying to prove the existence of an element which is indeed exist, but as you have seen, I thought that this operator had had a identity element. – 2017-01-07
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    I mean I also used the method of contradiction while proving that the previous operator is not associative. – 2017-01-07
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    @Leth Is $Q$ the set of rational numbers? If so, you're getting into some pretty nitty-gritty stuff that depends on how $Q$ is defined and what properties it is assumed to have (normally, we're OK freely using the fact that $0$ is the additive identity of the set of rational numbers), that's likely considerably more difficult than what you intended it to be. – 2017-01-07