Prove: There exists a real number $x$ such that for every real number $y$, we have $xy=y.$
In class I learned that I can prove a statement by:
- proving the contrapositive,
- proof by contradiction,
- or proof by cases.
Can I do something along the lines of $xy=y$
divide both sides by $x$ to get $y=\frac{y}{x}$
which would then be equal to $y=\frac{y}{x}=xy$
divide everything by $y$ to get $1=\frac{1}{x}=x$
so $x=1$?
There exists a real number $x$ such that for every real number $y$, we have $xy=x$
I know I somehow have to prove $x=0$ right? But I am not sure how to go about this.
Best Answer
Questions like this are difficult to give good answers to, not because the proofs themselves are difficult or deep, but because it's unclear what you're allowed to assume as already known, hence what constitutes an "acceptable" proof. It's tempting to say the proofs here boil down to saying "Let $x=1$" (for the first statement) and "Let $x=0$" (for the second), because all you have to show is the existence of a real number with a given property. Indeed, the first statement is, in essence, one of the axioms for the real numbers. The second statement isn't an axiom, but it follows from the theorem that $0\cdot y=0$ for all $y$, provided you have that theorem at your disposal; the existence of the number $x=0$ comes from the axiom for the additive identity.