# Math Help - What is the square root of 4?

1. ## What is the square root of 4?

square root of 4= ± 2
since 22 = 4
and (-2)2 = 4

is the above ture as we were taught that square root of 4=+2 only

2. Originally Posted by nazz
square root of 4= ± 2
since 22 = 4
and (-2)2 = 4

is the above ture as we were taught that square root of 4=+2 only
It is true that there are two square roots of 4, which are 2 and -2.

However, often we use "square root" to mean "principal square root" which is the nonnegative value, 2. With a radical sign $\sqrt{4}$ the value is by definition the principal square root. If we say "the square root" this is a pretty clear indication that we mean principal square root. Otherwise we would say "a square root." I think it is a slight abuse of terminology but doesn't really cause communication problems.

Square root - Wikipedia, the free encyclopedia

3. $\sqrt{4} = 2$ ... only

if $x^2 = 4$ , then $x = \pm 2$ ... there is a difference.

4. Originally Posted by nazz
square root of 4= ± 2
since 2(2) = 4
and (-2)2 = 4
is the above ture as we were taught that square root of 4=+2 only
Think of it this way: there are two square roots of 4, $\pm2$.

But the symbol $\sqrt{4}$ stands for only one number, $2$.

$-\sqrt{4}=-2$.

5. Originally Posted by Plato
Think of it this way: there are two square roots of 4, $\pm2$.

But the symbol $\sqrt{4}$ stands for only one number, $2$.

$-\sqrt{4}=-2$.
So symbol $\sqrt{-4}$ stands for $2i$ and $-2i$ is not included?

6. The reason we write the solution to $x^2= a$ as $\pm\sqrt{a}$ is that " $\sqrt{a}$" itself only stands for the "non-negative number such that x^2= a".

But your last question raises an ambiguity. The real numbers can be "ordered"- that is we can define "less than" and so "positive", "negative", and "non-negative". Since the imaginary numbers are just a real number times i, we can do the same and say that "ai" is "positive" if and only if a is positive. In that sense we could define "the" square root of -4 as "2i" and not "-2i".

However, the complex numbers, as a whole, cannot be ordered in such a way that the basic rules for an "ordered field"
a) For any two numbers a and b, one and only one holds:
i) a= b
ii) a< b
iii) b< a
b) If a< b then a+ c< b+ c for all c
c) If a< b then ac< bc for all c such that c< 0.

Because of that, for general complex numbers we cannot define "positive" or "non-negative" and have no consistent way to distinguish between the two roots of " $x^2= i$". That is why, generally, we must allow multiple valued functions and take either root of $x^2= a$ as " $\sqrt{a}$".

That is why there is an ambiguity here. We can either consider " $\sqrt{-a}$", for a a positive real number, to be "[tex]i\sqrt{a}[/itex]" and assert only that single value as the square root, or we can consider " $\sqrt{-a}$ to be the square root of a complex number and so take both values as the "square root".

Strictly speaking the same problem occurs with square roots of positive real numbers, because the real numbers are a subset of the complex numbers. That is, we can think of $\sqrt{4}$ as the square root of a real number and so have value 2 only or we can think of it as the square root of the complex number 4+ 0i and assert that both 2 and -2 are correct. But the real numbers have a "precedence", at least historically, so that we almost never do the latter.