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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?

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    $\begingroup$ The "if A then B" construction means, whenever A is true, B is required to be true, not merely an option. $\endgroup$ – Joffan Mar 23 '15 at 13:15
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    $\begingroup$ "If a basket contains $2$ apples or $1$ orange, then it contains $2$ apples" --- Is this statement true or false? $\endgroup$ – Santiago Canez Mar 23 '15 at 21:12
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    $\begingroup$ I note in passing that the context may matter here - what kind of a thing is $x$? The equation is equivalent (if additive inverses are present) to $x(x-1)=x^2-x=0$. Working in integers modulo $6$, where there are zero divisors, $x=3$ is a solution as well. Check $3^2=9\equiv 3 \bmod 6$ $\endgroup$ – Mark Bennet Mar 24 '15 at 10:04
  • $\begingroup$ If you are in a group and $1$ is the identity element, then the claim is true by the cancellation law of groups. In full generality the claim is false. (In full generality it is not even true that $1^2=1$, for instance if you are in a ring, where $1$ is the multiplicative neutral element and the operation you are considering is the "sum" operation). See my answer below for more details. $\endgroup$ – user126154 Mar 24 '15 at 10:08
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"$x = 1$" 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 "$x^2 = x$".

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    $\begingroup$ So it's incorrect to say "$x=1$ and $x=0$ are solutions"? $\endgroup$ – Ovi Nov 27 '15 at 9:38
  • $\begingroup$ @Ovi: Strictly speaking, it is incorrect. What you actually mean is that "$x=1$" and "$x=0$" are two possible solutions of the equation "$x^2=x$", which does not mean that both of them necessarily hold. And of course, they can't both hold at the same time. It is better to keep it simple and say "If $x^2 = x$, then either $x = 1$ or $x = 0$.", which is totally unambiguous and accurate. $\endgroup$ – user21820 Nov 27 '15 at 9:46
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It's false. We are not given enough information from $x^2=x$ to deduce that $x=1$.

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  • $\begingroup$ nice answer ...........+1 $\endgroup$ – Bhaskara-III Apr 23 '17 at 5:53
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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?

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  • $\begingroup$ yes, i can find it is $x=0 \quad (x\ne 1)$ for which $x^2=x$ $\endgroup$ – Bhaskara-III Apr 23 '17 at 5:40
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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?

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  • $\begingroup$ If the sneakers have the autographs of the cast of the movie Sneakers, and the candy is in fact diamonds coated with chocolate to smuggle them into the country, I think it could be a total amount of a million dollars or more. $\endgroup$ – Asaf Karagila Mar 24 '15 at 18:19
  • $\begingroup$ It could be! But this time, it wasn't! $\endgroup$ – MJD Mar 24 '15 at 18:54
  • $\begingroup$ I could believe you, but you already claimed to be a liar, so why would I? (On the other hand, if my suggestion is really the case, then you're not a liar and therefore your comment that it's not the case is probably true. We're a couple of regular Cretans, and also Godel sentences.) $\endgroup$ – Asaf Karagila Mar 24 '15 at 19:02
  • $\begingroup$ But I wasn't lying. Just ask nahtanoj! $\endgroup$ – MJD Mar 24 '15 at 19:21
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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 $0\ne 1$.

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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 $x=1$.

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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 $$x\cdot x=1\cdot x\Rightarrow x=1$$

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From $$p\implies q\vee r$$ you can not conclude $$p\implies q$$ For your example: $0^2=0$ but $0\ne 1$.

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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.

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