Number multiplied by itself does not give a square number

Question

The answer to this is probably very simple but while working on a question I was surprised to discover than a number multiplied by itself does not give the same answer as the same number squared (in the case of negative numbers).

Examples:

$-1^2 = -1$

If I now multiply this number by itself $-1 * -1$ I get $1$

$-4^2 = -16$

$-4 * -4 = 16$

The same goes for all the other negative numbers I tried.

What changes when squaring a number and how do I understand what is going on? When I multiply a positive number by itself I get a square number, why does this not work with negative numbers?

It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$. — Robert Israel, Mar 14 '19 at 16:49
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $x\cdot x$. — Nick Peterson, Mar 14 '19 at 16:50

score 0 · Accepted Answer · answered Mar 14 '19 at 16:53

0

Note that the square of $-1$ is $$ (-1)^2 = (-1)\times (-1) = 1 \ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) \times (1) = - 1 . $$

In general $-(x)^2$ is $- (x\cdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.

answered Mar 14 '19 at 16:53

Dando18

5,368

I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected – Xantium Mar 14 '19 at 16:59

Number multiplied by itself does not give a square number

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