If $x^2=x$, then $x=1$. Is this statement true or false? I understand that this equation has two solutions, that is $x=1$ or $x=0$. But if you say this statement is false, you are like saying that $x=1$ is not the solution for the equation. If a basket contains $2$ apples and $1$ orange, and I say that the basket contains $2$ apples. Is there anything wrong? Anyone have a good argument one way or the other?
 A: It's false. We are not given enough information from $x^2=x$ to deduce that $x=1$.
A: It's better to look at the contraposition. 
Think about it: if $x \neq 1$, does it necessarily follow that $x^2 \neq x$? Can't you find another number which satisfies the equation?
A: The answer may be easier to see if we use quantifiers. Your original statement would then be:
$\forall x\in\mathbb{R}:[x^2 = x\implies x=1]$
Or equivalently:
$\neg \exists x\in \mathbb{R}:[x^2 =x\land x\ne 1]$
Clearly this is false since, for $x=0$, we have $0^2=0$ and $0\ne 1$.
A: Suppose I state “If you bring me the golden idol of Horus, then I will pay you one million dollars.” So you risk your life to get the idol, and you bring it to me.  Then I give you a pair of sneakers and a bag of candy.  You say “Where's my million dollars?”  I say “Well, I could have given you a million dollars.  It just happens that I didn't.”
Was my statement true or false?
A: Truth is not universal as we would like to think. It is relative to the context.
In the real numbers, or the integers, or something similar where $0^2=0$ and $1^2=1$, the statement is false. Because there is a counterexample, $0^2$. 
If the context is something like $\Bbb N$, and $0\notin\Bbb N$ (recall that in some contexts $0$ is a natural number, and in other contexts it is not), then the statement is true. Because in the positive natural number, there is a unique solution for $x^2=x$ and that is $x=1$.
A: A part from the logical problem of reversing implications:
$$x=1\Rightarrow x^2=x\qquad \text{is different from}\qquad x^2=x\Rightarrow x=1$$ 
If you don't specify the nature of $x$, than the claim is largely false in general, as it is not true that $x^2=x\Rightarrow x=0,1$. On the other hand, there are algebraic structures where the claim is true.
Example. 
Let $f:\mathbb R^3\to\mathbb R^2$ be the projection on the $XY$-plane along the $Z$-axis.
Then, for every $p\in\mathbb R^3$ you have $f(f(p))=f(p)$. Thus $f^2=f$. Nonetheless $f$ is different from both $1$ (the identity) and $0$ (the zero map.)
When the claim it is true:
If you are in a Group, where there is a unique identity and all elements are invertible, then by the cancellation law (that say $ax=bx\Rightarrow a=b$) you have $$x\cdot x=1\cdot x\Rightarrow x=1$$ 
A: "$x = 1$" is not at all the same as "$x$ can be $1$". The former states in no uncertain terms that $x$ can only be $1$ and nothing else, but that is not true if you are just given "$x^2 = x$".
A: From $$p\implies q\vee r$$ you can not conclude $$p\implies q$$
For your example: $0^2=0$ but $0\ne 1$.
A: The difference between the fruits thing and $x^2=x$ is that $x^2=x$ is equivalent to $x = 0$ or $x = 1$. In your fruits thing, you said 2 apples AND 1 orange.
