Give situations to show that square root of a^2 is not equal to a. And use this to show that square root of a^2 is equal to the absolute value of a.

Note by Noel Quirol
4 years, 10 months ago

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1. Absolute value of a real number is it's distance from 0. If you have a point on the real line, call that point P. Let the origin be O, imagine a square with its base PO. What is the length of this square?
2. Induction is for natural numbers but I think there is this paper on using induction argument on statements defined on the set of real numbers.

- 4 years, 10 months ago

I think it has something to do with induction ? Am I right ? But I dont know where to start. ..

- 4 years, 10 months ago

I can't see how your problem has anything to do with induction.

Start with the definition of square roots. From what I understand, you're supposed to show examples. That shouldn't be tough.

A hint that makes it really easy:

What is $$\sqrt{(-4)^2}$$?

- 4 years, 10 months ago

Comment deleted Jul 28, 2013

Okay, this is going to be a little harder than I thought.

Take a look at this: $$\sqrt{x }$$

Do you know what this symbol around $$x$$ is called? It is called the principal square root operator or sometimes simply the square root operator. Whenever people are talking about the square root of a number, they are basically talking about the principal square root.

It is true that $$16$$ has two square roots. One of them is $$\sqrt {16}$$. The other one is $$-\sqrt{16}$$. If you square both of these numbers, you are going to get $$16$$.

But the principal square root of $$16$$ is just $$\sqrt{16}=4$$, not $$-4$$.

So the square root (principal) of $$(-4)^2$$ or $$16$$ is just $$4$$.

I think you can take it from here.

- 4 years, 10 months ago

thanksss

- 4 years, 10 months ago