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Can I have an answer please ??

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, 2 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.

Okay Nho - 4 years, 2 months ago

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

Noel Quirol - 4 years, 2 months ago

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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 \(\sqrt{(-4)^2}\)?

Mursalin Habib - 4 years, 2 months ago

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Comment deleted Jul 28, 2013

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@Noel Quirol 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.

Mursalin Habib - 4 years, 2 months ago

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@Mursalin Habib thanksss

Noel Quirol - 4 years, 2 months ago

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