# What would happen if $\sqrt{}$ instead meant the negative square root?

I read that it is only by convention that $\sqrt{}$ means “the positive square root of”. The downvotes and user 'Thomas' 's answer compel me to clarify that I asked this hypothetical question only because of curiosity, not because of any desire to desecrate $\sqrt{}$.

What if $\sqrt{}$ meant “the NEGATIVE square root of”? Why might convention not have caused this?

Source: Page A7, Appendix A, Calculus Early Transcendentals (6th ed.; 2008) by James Stewart:

Recall that the symbol $\sqrt{}$ means “the positive square root of.”
Thus $\sqrt{r} = s$ means $s^2 = r$ and $s \ge 0$.
Therefore, $\color{darkred} { \text { the equation$\sqrt{a^2} = a$is not always true. It is true only when$a \ge 0$} }$.
If $a < 0$, then $-a > 0$, so we have $\sqrt{a^2} = -a$.
[...] we then have [...] $\sqrt{a^2} = |a|$.

• It would just be an annoying change in notation. I'd have to start writing $- \sqrt{a}$ instead of $\sqrt{a}$ for no reason. – Qiaochu Yuan Oct 30 '15 at 20:30
• There is already an operator that does this. It's denoted $-\sqrt{}$. – David Hill Oct 30 '15 at 20:31
• We would no longer have $\sqrt{xy}=\sqrt x\sqrt y$, for example (take $x=y=2$). This seems like a desirable property to me. – Clayton Oct 30 '15 at 20:31
• This is just one of the 2 branches of $\exp{\frac{1}{2} Log z}$. It is only a convinient way to denote The positive number $a$ that satisfies $a^2 = x$. – Ranc Oct 30 '15 at 20:40

You are free to change notation. The only thing that is going to happen is that it will confuse and annoy a lot of people. Also, if you use notation that is contrary to established notation without giving a good reason, people aren't going to take your serious as a mathematician.

As mentioned in the comments above, if you define $\sqrt{a}$ to be the negative number $b$ such that $b^2 = a$, then we would just have to put a minus in front of all square roots appearing in the literature. It would (as also mentioned in the comments above) cause the problem that $$\sqrt{ab} \neq \sqrt{a}\sqrt{b}$$ And then you have have to make sense of stuff like $x^{1/3}$. How is this now defined? There are good reasons for picking the notation we have.

So basically, you have to ask yourself: why would you want to do that?

• +1. Thank you, but your first paragraph sounds critical of my question, which I just elucidated. Does my emendment help? – Greek - Area 51 Proposal Oct 30 '15 at 20:48
• @LePressentiment: Sorry if I sounded critical of your question. I decided to answer it because it does make sense to ask. The "What if" part of the question is basically answered by: nothing would happen except you would need to replace all $\sqrt{}$ by $-\sqrt{}$ in expressions. You would also not have various formulas true anymore. – Thomas Oct 30 '15 at 23:17
• No problem; thank you for your elucidation! I just wished to confirm my respect for $\sqrt{}$. – Greek - Area 51 Proposal Nov 1 '15 at 0:35

So imagine that convention caused $\sqrt{}$ to mean “the NEGATIVE square root of”. Then what would change or differ?

We would get negative signs everywhere instead of positive ones. This will happen every-time we solve an equation with even powers.

Would mathematics be fundamentally changed?

Good question, but I don't think it would, only the signs will change.

As commenters said, mathematics would be the same up to the substitution $\sqrt{\ } \mapsto -\sqrt{\ }$.

Actually, a similar phenomenon already happens in the context of differential geometry and PDEs. The same symbol $\Delta$ means $-\frac{\partial^2}{\partial x_1^2} -\ldots -\frac{\partial^2}{\partial x_n^2}$ in the geometers's literature and $\frac{\partial^2}{\partial x_1^2} +\ldots +\frac{\partial^2}{\partial x_n^2}$ in the PDEs literature. It's a bit confusing but people are used to it.