Prove there exists a real number $x$ such that $xy=y$ for all real $y$ 
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.
 A: As it turns out, in the first part, you managed to find the correct $x$ along the way, but not very efficiently, and you overlooked a significant issue! Let's see how we can go about finding it deliberately, without issue.
If it doesn't immediately occur to us what value of $x$ might allow $xy=y$ to be true for all real $y,$ then we can start by trying to deduce a candidate for $x,$ then check to see if said candidate actually works for all real $y.$ Let's look at a couple of possible approaches to this, in order from least useful to most useful (in general).

Method 1: Suppose that $xy=y$ for all real $y.$ That means we can pick any particular real $y$ we want, and the equation must hold. Let's try $y=7,$ say. Then $xy=y$ becomes $7x=7,$ which we can easily solve to get $x=1.$ In fact, since this is the only solution $x$ to $xy=y$ when $y=7,$ then it is the only solution that might make $xy=y$ for all real $y.$ It is readily checked that if $x=1,$ then $xy=1y=y,$ and so we're done.
The drawback to this approach is that it relies to some extent on a good choice for $y.$ In this particular problem, that isn't a big deal, but in many problems it won't be so simple.

Method 2: Again, suppose that $xy=y$ holds for all real $y.$ We'd like to solve for $x.$ Now, it's tempting to divide both sides by $y$ to isolate $x$ on the left-hand side, but wait just a moment! There is a real number that we cannot divide by! In particular, we cannot divide by $y=0.$ Fortunately, if $y=0,$ it doesn't matter what $x$ we choose--the equation $xy=y$ will hold automatically, since both sides are $0.$ Whew! That issue has been averted. But what if $y\ne 0?$ Well, in that case, we can divide both sides by $y,$ yielding (as you saw in your work) $x=1.$ Again, this is easily confirmed to be correct.
The drawback to this approach is that we have to split by cases, and we have to notice that we have to split by cases (or risk division by $0$). Note that this approach uses the fact that $0y=0$ for all real $y.$ If you've not seen this result proved, I recommend you try to prove it, using the fact that $0+0=0,$ so that $(0+0)y=0y.$ See if you can take it from there.

Method 3: Again, suppose that $xy=y$ for all real $y.$ Put another way, $xy-y=0$ for all real $y,$ or by factoring, $(x-1)y=0$ for all real $y.$ Now, the only way two real numbers can have a product of $0$ is if at least one of the numbers is $0,$ so we know that $x-1=0$ or $y=0.$ (Note: This includes the possibility that both are $0.$) Since $y$ could be any real number, then we can't assume that $y=0,$ and so we must have $x-1=0,$ from which we once again find that $x=1.$
This approach is the best in terms of low risk, fewest cases, and general applicability. However, you may or may not have it to work with. I recommend that you attempt to prove it, if you haven't seen it proved. That is, prove the following statement:

If $a,b$ are real numbers such that $ab=0,$ then $a=0$ or $b=0.$


As for the second problem, we can use one of the above sorts of approaches to discover what $x$ must be (and I recommend that you practice the third method in particular to verify this), but you've already intuited that $x$ must be $0$! All you have to do now is observe that if $x=0,$ then for any real $y,$ we have $xy=0y=0=x,$ and so you're done!
A: 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.
A: Your approach, showing that $x=1$, works, but you could make it a lot simpler by dividing by $y$ to begin with. Then you get $x=1$ as a result of your first step.
For both of these, remember that you are proving that such a value exists; if you find a specific example, you don't have to explain how you found it (unless an assignment says otherwise). You aren't trying to prove that $xy=x\Rightarrow x=0$. You're proving that there is a value of $x$ where $xy=x$ is true for all $y$. If you can find that value, then your proof is simply showing that the value fits the conditions.
As you observe in the comments, you can simply say "Let $x=0$. Then $xy=0\cdot y=0=x$."
A: Since $\mathbb R \setminus \{0\}$ is a group (wrt. multiplication), the identity satisfies this equation. The identity is clearly $1$.
A: There exists a real number $x$ such that for every real number $y$, we have $xy=y$. ? If such a number $x$ exists then it should be the number $1$. In fact, since $xy=y$ for any $y$ then one should have $xy=y$ when $y=1$ then $x.1=1$ that is $x=1$. Now, If $x=1$ is it true that $xy=y$ for every $y$? the answer is yes since in this case $xy=y$ is nothing than $y=y$ which is evendently true for every $y$. We have then proved that $x=1$ is a necessary condition and next we have proved that it is a sufficient condition to have $xy=y$ for every $y$. Q.E.D
A: Note that in almost any algebraic structure - at least one which is distributive with a multiplicative identity (which we call $1$) you will find that  $xy=y$ implies that $xy-y=0$ or $(x-1)y=0$ from which you can conclude (if there are no zero-divisors other than $0$) that $x=1$ or $y=0$.
Since you need a value to work for all $y$ the solution is $x=1$
If you needed a value for all $x$ it would be $y=0$
A: 
I'm not really sure how to get started on these problems. In class I learned that I can prove a statement by: proving the contrapositive proof by contradiction or proof by cases

In your position, I wouldn't be sure how to get started either, because it sounds like you haven't been given a starting point.
To prove something, you need two things:


*

*A set of basic assumptions which do not need to be proven.

*A set of reasoning techniques you can use to show that the assumptions imply the desired conclusion.


It sounds like you've been given some reasoning techniques (proving the contrapositive, proof by contradiction, and proof by cases), but you haven't been given any basic assumptions to start from, so you can't prove anything! You should ask your teacher which basic assumptions you should use.
In technical writing, the basic assumptions and reasoning techniques are usually called axioms and rules of inference, respectively.
Many mathematicians describe the real numbers using a set of basic assumptions called the complete ordered field axioms. As @BarryCipra mentioned, this set of axioms actually includes the first statement you're trying to prove—so if you use the complete ordered field axioms as your starting point, your proof of the first statement will be really short!
Of course, there are many other sets of axioms you can use to describe the real numbers. One of my favorites includes some assumptions that might be phrased, in very informal language, like this:


*

*Every rational number is a real number.

*Every real number can be approximated by rational numbers to arbitrarily high precision.

*As you make a real number $a$ closer and closer to $a'$, and $b$ closer and closer to $b'$, the sum $a + b$ gets closer and closer to $a' + b'$, and the product $ab$ gets closer and closer to $a' b'$.


You might enjoy thinking about how you could use assumptions like these, restated in a suitably precise way, to prove what you want to prove. (You'll need some additional assumptions I didn't mention.)
A: I am pretty sure that you can just pick x=1 in the real numbers. so for every real number y there exists xy=y because x=1 implies y=y which is true for every real number.
