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Is $-2^2$ equal to $-4$, or to $4$?

Mathematics Asked by tmakino on January 1, 2022

Is it always safe to assume that $-2^2 = -4$, or is this dependent on notation, i.e. would it be mathematically correct (albeit sloppy) to say $-2^2 = (-2)^2 = 4$? Specifically, I’m asking in the context of the GRE General Test, if anyone is familiar with the particulars of their notation.

3 Answers

The conventional interpretation of the notation is that $-2^2 = -(2^2) = -4$.


The order of operations

I think that the other answers are wrong in asserting that $-2^2 = -4$ because of the order of operations, or that it follows from the order of operations. The order of operations is an entirely arbitrary set of conventions for how to read and understand mathematics. This order of operations is not a mathematical result which can be proved. Rather, it is a set of rules which we, as practicing mathematicians, have agreed upon.

These rules are roughly as follows:

  1. If a group of symbols is enclosed in a pair of (generally matching) brackets or grouping symbols, e.g. $()$, ${}$, $[]$, $lfloor rfloor$, $||$, $||$, etc., then that group of symbols should be dealt with before anything from outside the group has a chance to act. In other words, highest priority is given to groups which the author has intentionally put together with some kind of grouping symbol.

  2. Inside of a set of grouped symbols (or if there are no grouping symbols), exponentiation take precedence over other operations. Exponentiation is handled from right-to-left. For example, $$ 4 mathrel{hat{}} 3 mathrel{hat{}} 2 = 4^{3^{2}} = 4^{(3^2)} = 4^9 = 262144. $$ The reason for the seemingly-contradictory right-to-left interpretation is that exponentiation has the property that $(a^b)^c = a^{bc}$, hence if we mean $(a^b)^c$, then it is just as easy to write $a^{bc}$; thus we can unambiguously use $a^{b^c}$ for $a^{b^c}$). The precedence of exponentiation over other operations will be justified / rationalized below.

  3. Multiplication is the operation with the next highest precedence. Multiplication is commutative, so the order in which multiplications are handled is irrelevant. I am intentionally not listing multiplication and division together, as I think that this causes confusion: eliminate division from your thinking entirely. In the expression $1div 3$, we should immediately interpret the symbols $div 3$ as $times frac{1}{3}$. Thus, after we change all divisions to multiplications, the order doesn't matter.

    Note, also, that this way of thinking immediately clears up a common misconception: many students are taught some kind of mnemonic, such as PEMDAS or BODMAS, where it might appear that multiplication should be given priority over division (or vice versa). If we understand that division "is just" multiplication by the reciprocal, there is no issue.

  4. Addition is done last. Again, addition is commutative, so order doesn't matter. In the same way that division "is just" a kind of multiplication, subtraction "is just" addition of the additive inverse. Make the substitution $-a$ is $+(-a)$, and add.

If you need a mnemonic for this, I like GEMA, for groups, exponents, multiplication, and addition.

Why this order?

Knowing that there is an order of operations is different from understanding why it is the way that it is. I think that the most compelling argument is that the usual order of operations is the convention which let's us most easily and unambiguously write out polynomial expressions in their "standard" form. This form is useful because it allows us to very easily verify if two polynomials are equal (all we have to do is compare coefficients), and certain other operations (such as addition) are done more efficiently.

With no order of operations, we would be forced to write things like $$ [ acdot (x mathrel{hat{}}3) ] + [b cdot (x mathrel{hat{}} 2) ] + [ccdot x] + d qquadtext{or}qquad [a cdot (x^3) ] + [b cdot (x^2)] + [c cdot x] + d. $$ These are quite clunky. Because polynomials are such a fundamental part of so much of mathematics, we really want a convention which is going to eliminate this clunkiness. Via the usual convention, the above expressions can be written more simply as $$ ax^3 + bx^2 + cx + d, $$ which is much cleaner and easier to read.

Back to the beginning

Getting back to the original question: what is $-2^2$?

Because we want polynomials to have simple, efficient representations, we have adopted the custom that exponentiation should be handled before any kind of multiplication or addition. As such, we deal with the exponent before anything else. Hence $$ -2^2 = -(2^2) = -4. $$ It might help to think about this expression as part of a larger expression (I've highlighted the expression of interest in red): $$ 9 color{red}{- 2^2} = 9 color{red}{- 4} = 9 + (color{red}{-4}) = 5. $$

Answered by Xander Henderson on January 1, 2022

No, in general it is said that $-x^2 implies -(x^2)$. This follows from BIDMAS (PEDMAS, BODMAS etc $ad nauseam$), since first the indices are evaluated, then the sign (which comes under multiplication, since $-x=-1 times x$) is given.

Answered by aidangallagher4 on January 1, 2022

This is order of operations. If you write $$ -2^2=-1*2^2 $$ pemdas dictates we do the exponentiation first. So no, this is not the same as $$ (-2)^2 $$ which privileges the multiplication by putting it in parentheses.

Answered by operatorerror on January 1, 2022

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