Confusion about notation of negative numbers

Question

The convention is that $-3^2 = -9$. But I think this implies that "-3" does not refer to the integer that is a solution to $3+x=$0. Otherwise $-3^2$ would mean "$-3$ times $-3$", the same as $(-3)^2$.

Secondly, I used the following logic to prove that $-3^2$ is in fact the same as $(-3)^2$:

$(-3) = -3$ and $-(3) = -3 \implies -(3) = (-3)$ which implies that $(-3)^2 = -(3)^2$. But this is incorrect.

What am I missing? And more importantly, what does "$-3$" mean and why does it seem like brackets change what "$-3$" means?

Edit: Seems like I have been bad at explaining what I want to know. My problem here is that there is apparently no way to write a negative number as one unit (for a lack of a better term). "$-n$" does not mean a number directly, but rather the number that is the result of a unary operation and a number $n$. So why is this the way it is? Why doesn't our notation allow us to directly write a negative number as one unit?

I literally started the post by saying that $-3^2 =-9$ so the answers pointing to a supposed duplicate is not helping.

Answer 1

The expression $-3^2$ involves two operations:

• Negation, aka “additive inverse” or “unary minus”. (Although it uses the same $-$ symbol, this is a distinct operation from subtraction, which takes two operands.)
• Exponentiation. (When the exponent is 2, this is given the special name “squaring”.)

The question is, which one takes precedence?

• If negation has higher precedence, then $-3^2 = (-3)^2 = 9$.
• If exponentiation has higher precedence, then $-3^2 = -(3^2) = -9$.

The standard convention is that exponentiation takes higher precedence than negation. A nice feature of this convention is that it simplifies writing polynomials with negative leading coefficients, e.g., $- x^2 + 6x - 9$. If negation had precedence, then we'd have to write $-(x^2) + 6x - 9$ instead.

Thus, $-3^2 = -9$.

Answer 2

The mistake in your logic is where you say that “$-(3)=(-3)$ implies that $(-3)^2 = -(3)^2$.” If two things are equal, their squares are equal, but the square of $-(3)$ is not $-(3)^2$. The square of $-(3)$ is $(-(3))\cdot(-(3))$, which is different from $-3^2$, because exponentiation has higher precedence than the unary minus operator.