Square Root: Positive & Negative

[mathdad]

Why do we get two answers when taking the square root?

For example, let a = any positive number

sqrt{a} = - a and a.

Why is this the case?

What about 0?

Can we say sqrt{0} = - 0 and 0?

 

[Evgeny.Makarov]

Why do we get two answers when taking the square root?

By definition. Wikipedia says that a square root of a number $a$ is a number $y$ such that $y^2$ = $a$. If $y$ is a square root of $a$ according to this definition, then so is $-y$ since $(-y)^2=((-1)y)^2=(-1)^2\cdot y^2=y^2$. It is probably slightly more difficult to explain why there are at most two square roots. The notation $\sqrt{a}$ denotes the principal square root, which by definition is a nonnegative square root.

What about 0?

Can we say sqrt{0} = - 0 and 0?

We can, but $0$ and $-0$ is the same number.

 

[mathdad]

By definition. Wikipedia says that a square root of a number $a$ is a number $y$ such that $y^2$ = $a$. If $y$ is a square root of $a$ according to this definition, then so is $-y$ since $(-y)^2=((-1)y)^2=(-1)^2\cdot y^2=y^2$. It is probably slightly more difficult to explain why there are at most two square roots. The notation $\sqrt{a}$ denotes the principal square root, which by definition is a nonnegative square root.

We can, but $0$ and $-0$ is the same number.

Excellent. Good job!

 

[HallsofIvy]

Why do we get two answers when taking the square root?

For example, let a = any positive number

sqrt{a} = - a and a.

No. A number can have two square roots but the two square roots of a are not "a" and "-a", they are "$$\sqrt{a}$$" and $$-\sqrt{a}$$" where $$\sqrt{a}$$ is the squae root I referred to before.

Why is this the case?

Because $$(\sqrt{a})^2= a$$ by the definition of "square root" and $$(-\sqrt{x})^2= (-1)^2(\sqrt{a})^2= (1)(a)= a$$.

What about 0?

Can we say sqrt{0} = - 0 and 0?

 

[mathdad]

Thank you everyone.

 

[HallsofIvy]

And, if we allow complex numbers, then the equation $$x^3= a$$. where a can be any complex number, has three solutions, $$x^4= x$$ has four solutions, and, in general, $$x^n= a$$ has n solutions.

If we restrict ourselves to real numbers, then the equation $$x^n= a$$, for a any real number and n odd, has one solution, while $$x^n= a$$, for n even, has 0 solutions if a<0, 1 solution if a= 0, and 2 solutions if a> 0.

 

[mathdad]

And, if we allow complex numbers, then the equation $$x^3= a$$. where a can be any complex number, has three solutions, $$x^4= x$$ has four solutions, and, in general, $$x^n= a$$ has n solutions.

If we restrict ourselves to real numbers, then the equation $$x^n= a$$, for a any real number and n odd, has one solution, while $$x^n= a$$, for n even, has 0 solutions if a<0, 1 solution if a= 0, and 2 solutions if a> 0.

Very useful.