Solving the equation $a+e+2ae=a$ for $w$

Question

Just need a quick answer of how my tutor got $e$ to $= 0$ from this equation. (I'm trying to find the identity of a binary operation)

$$a+e+2ae=a$$

I feel like this is a very easy problem but I'm grinding out some study right now and very tired.

Answer 1

It seems that you're trying to determine the identity of a binary operation $$a\star b:=a+b+2ab.$$ That is, you are trying to find some $e$ such that for every $a,$ we have $a\star e=a.$ Using the definition of $a\star e,$ we have $$a+e+2ae=a.$$ Subtracting $a$ from both sides and factoring out the $e,$ we have $$e(1+2a)=0.$$ At this point, context becomes very important. Assuming that there is at least one $a$ for which $1+2a\ne 0,$ and assuming that $xy\ne 0$ whenever $x\ne 0$ and $y\ne 0,$ it follows that we must have $e=0.$ Do you see why? Consider those two assumptions I mentioned, and let me know if you aren't getting anywhere.

Answer 2

$$a+e+2ae=a\underbrace{\Leftrightarrow}_{-a \, \mathrm{both}\,\mathrm{terms}}e+2ae=0 \underbrace{\Leftrightarrow}_{\mathrm{common}\,\mathrm{factor}} e(1+2a)=0 \\ \underbrace{\Leftrightarrow}_{e\ne 0}1+2a=0 \Leftrightarrow a=-\frac 12. $$