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I understand order of operations plays a large role, but I am still confused. I figured out:

If $-8^2 = -(8^2)$ then $-8^2 = -64$. Now the can do the inverse. To do the inverse we square root the entire expression. So $\sqrt{-64} = \pm8i$, or we can separate $-64$ into $-1 \cdot 64$ and we also get $\sqrt{-1}\cdot\sqrt{64} = i \cdot ±8 = \pm8i$. This means that we end up with different value to what we started with, meaning the inverse does not work. Now where $-8^2 = (8\cdot-1)^2$, we can say that $-8^2 = 64$, and doing the inverse we get $\sqrt{64} = \pm8$, which works.

So firstly, is there anything wrong with this proof. And secondly, if the proof above is valid, why is $-n^2$ seen as $-(n^2)$?

Matthew Towers
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MidnightAmethyst785
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4 Answers4

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Because $-8^2 = -(8^2)$ is the way that needs the fewest parentheses in the long run.

Order-of-operations is a man-made concept. We could very well do math entirely without it, or with a different order, and no objective part of math would be any different. The only thing that would change is writability and readability. We humans have decided that that's what it should mean because it's the most convenient, so that's what it means.

Arthur
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The inverse does work, but you're using the inverse of the wrong function. If your function is $x \mapsto -x^2$, the inverse relation is $x \mapsto \pm\sqrt{-x}$: in other words, $y = -x^2 \iff x = \pm\sqrt{-y}$. You can check that with $x = 8$, there is no contradiction.

Prasiortle
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The order of operations is just a convention, and over the years there have been attempts to introduce other conventions. Polish notation is an example.

The order of operations attempts to model the spoken language symbolically. With $-n^2$ being the negative of $n$ squared rather than the square of negative $n.$ In part this is because the $-$ symbol has double duty. It could indicate that we are negating a number or it could mean that we are subtracting a term. Since we need to do both, which interpretation will be more useful? If we want to read $-n^2$ as $(-n)^2$ then if we are subtracting the term $n^2$ we will need to write $-(n^2).$

We have decided to read the exponentiation as happening before the subtraction or negation. But, again this is a convention. One that seems to be useful most of the time.

As for your algebra. If we want to invert the function $y = -n^2$ then $n = \pm \sqrt {-y}$

user317176
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$-a$ is generally an abbreviation for the solution of $x+a=0.$ This means, that $-a$ stands for the inverse element of an additive written group. It is a bit of a sloppy notation since the minus sign abbreviates what would be $ a^{-1}$ in the multiplicative case.

Thus, $-n^2$ is a solution for $x+n^2=0.$ The question, whether it should represent $(-n)^2$ instead doesn't even occur. BEDMAS and alike has no say here either. $-a$ is short for $+(-a)$ regardless of what $a$ is or whether it is used in an equation or as a standalone.

Marius S.L.
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    I don't like this answer, because you could easily turn it on its head. "$a^2$ is generally an abbreviation for $a\cdot a$. This means, that $a^2$ stands for the result of multiplying $a$ with itself in the multiplicative written monoid. Thus, $-n^2$ is the result of multiplying $-n$ with itself. The question, whether it should represent $-(n^2)$ instead doesn't even occur." There is no mathematical reason why your answer is more correct than what I have put in quotes here. Only human conventions. – Arthur Jan 01 '24 at 20:22
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    @Arthur And I do not like the minus sign in any other context. It is a source of confusion and "law by authority". The minus sign is ambiguous if you do not read it as what it is, the inverse group operation: $+a^{-1}.$ If you read it as such, 90% of misconceptions are immediately gone. – Marius S.L. Jan 01 '24 at 20:27
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    This completely misses the point. Even if $-n^2$ did mean $(-n)^2$, the minus sign still denotes the additive inverse with respect to $+$. It is just the one of $n$, not the one of $n^2$. Also, it is not true in general that adding a minus to the front of any expression gives its negative. E.g. $-1 + 1$ is not the negative of $1 + 1$. – Tzimmo Jan 01 '24 at 21:27
  • @Tzimmo This is not what I said! I never claimed it meant $(-n)^2$. Please do not quote me wrong! All I said is, that $-a\in (G,+)$ is the same as $a^{-1}\in (G,\cdot).$ It is neither a binary operation nor is it the "opposite" of "+". It is a bad choice of notation. The correct notation would be $-a:=+(-a)$ or $-a:=+(a^{-1}).$ That is what I really said. I admit, that Kurosh has left his traces in my understanding and that it is not how it is taught. But that is the typical teacher attitude: teach it wrong and keep the sovereignty of interpretation. Sad, but it is as it is. – Marius S.L. Jan 01 '24 at 22:05
  • Now you are quoting me wrong. I never said that that was your claim. My point is that your argument works not only for the correct interpretation, but also for the wrong one. It does not help in distinguishing both. Moreover, I do not see any upside in the alternative notations you propose for the negative. They both overload the $+$-sign, the second also overloads $\bullet^{-1}$, which is already used for a very different operation. – Tzimmo Jan 01 '24 at 22:22
  • I believe that both operations belong primarily to a group or a semi-group in the case of rings. They only connect by the distributive law and any other combination is a misconception. If people would share this point of view then all this nonsense like BEDMAS would immediately vanish. $-a^2=(-1)\cdot (a\cdot a)$ where $-1$ is the solution to $x+1=0.$ That's it. Everything else is an interpretation intended to keep the sovereignty of judgment. All we have is the distributive law and two binary operations. – Marius S.L. Jan 01 '24 at 22:30
  • What does the unary $+$ in $+(-a)$ mean? – peterwhy Jan 01 '24 at 23:30
  • @peterwhy It means the binary plus as in $ \ast +(-a)$ or the unary long version of the informal "additive group inverse". The minus sign is a bad choice in my opinion mainly because it fails to be associative. You need some extra interpretations to use it. That disqualifies it in my opinion. Plus is well-defined, minus is a (in my eyes bad) convention. Strangely, nobody has problems accepting $":a"$ as $"\times a^{-1}"$ as its definition. I only want the same right for $-a= +\operatorname{inv}(a)=+(-a).$ – Marius S.L. Jan 01 '24 at 23:36
  • So does the unary $+$ mean identity operation here? Saying "$-a$ is short for $+(-a)$" here seems incomplete, as the $-a$ inside $+(-a)$ still needs a definition. (Is $+(-a)$ short for $+(+(-a))$? What does that mean again?) – peterwhy Jan 02 '24 at 00:27
  • If we have $a-b$ then it should be read as $a+(-b).$ This solves the problem with the associative law and reduces $-b$ to the element that solves the equation $x+b=0.$ I only left out "a" so that $a-b$ became $-b$ and $a+(-b)$ became $+(-b).$ It was the binary meaning. You can also read it as "+(-a) means the additive (+) inverse (-) element (a)." I simply do not see why we should treat addition and multiplication differently. Minus as a binary operation isn't associative, and minus as a unary operation only means a solution of the equation $x+(-x)=0$ that defines $-x.$ – Marius S.L. Jan 02 '24 at 00:41