The roots of, let's say 4, can be either 2 or -2. Why is this an inclusive-or i.e. if $x^2 = 4$, then $x = 2 \vee x = −2$, instead of an exclusive-or i.e. if $x^2 = 4$, then $x = 2 \oplus x = −2$? It can only be one of them because we can only get 4 if we do a $2 \times 2$ or a $-2 \times -2$. From my understanding, something that can only be one or the other, should be treated as an exclusive-or.

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It is an exclusive or. Why do you say it is an inclusive or? If it was an inclusive or there would be the possibility that $x$ equals two different values at the same time. – Shahab Sep 26 '12 at 12:37
@Shahab Ah, because when I answered a quiz in coursera.org/course/maththink, I said that it should be an exclusive-or because, as you said, there is no possibility that $x$ equals two different values. But I got the wrong answer and I'm not quite sure why. – Jag Talon Sep 26 '12 at 12:44

First of all, the notation $x = (2 \vee -2)$ does not make any sense. Do you mean $x = 2 \vee x = -2$? Also, note that the square root $\sqrt{a}$ is defined to be the positive solution of $x^2 =a$, so $\sqrt 4 \neq -2$.

Now while it is true that $x = 2$ and $x = -2$ can never be true at the same time, the situation changes if we have a quadratic equation with a double solution, e.g., $x^2 = 0$. This equation has solutions $x = 0$ and $x = -0$, but since $0 = -0$, $x = 0 \oplus x = -0$ does not hold. This is why it is preferable to use an inclusive or.

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Thanks for mentioning that case! – Jag Talon Sep 26 '12 at 12:56

This confusion would all cease if we referred to the $\sqrt{\,\cdot\,}$ symbol as 'flobble' rather than 'square root', since a positive real number $a$ has two square roots and $\sqrt{a}$ refers to only one of them. (The other is $-\sqrt{a}$.)

If $x=\sqrt{4}$ then $x=2$, not $-2$, since $\sqrt{4}$ refers exclusively* to the positive square root.

I'm not sure what $x = (2 \oplus -2)$ is meant to mean. If it means $x=2$ xor $x=-2$ as Shahab suggests in the comments, then... well let me explain. It's true that if $x^2 = 4$ then ($x = 2$ xor $x=-2$); but it is not true that $\sqrt{4} = 2$ xor $-2$. $\sqrt{4}$ does not refer to 'a' solution to $x^2=4$, it refers to 'the positive' solution to that equation.

*Except in certain contexts, but I don't think this is one of them.

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I think by $x = (2 \oplus -2)$ the OP meant the "exclusive or", i.e. 2 or -2 but not possibly both. – Shahab Sep 26 '12 at 12:25
Sorry about the notation. I modified my post and it should be clearer now. Thanks! – Jag Talon Sep 26 '12 at 12:47
@JagTalon: Okay sure. In that case my answer still applies. It is true that solving $x^2 = 4$ gives rise to an xor, since $x$ cannot take two values; but $\sqrt{4}$ takes just one value $-$ it is unambiguously equal to $2$. – Clive Newstead Sep 26 '12 at 12:49
@CliveN. Thanks for your answer! But what about the case where $x^2 = 0$ as Johannes mentioned? Would that make an inclusive-or? – Jag Talon Sep 26 '12 at 12:55
@JagTalon: What do you think? ;) – Clive Newstead Sep 26 '12 at 13:04