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## #1 2013-08-21 07:36:38

Al-Allo
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### Question on the square rooth of polynomials

Hi : http://img841.imageshack.us/img841/9800/hviw.png

I was wondering, I undestand what the text in the given link says, but I don't understand why we don't consider square rooth of variables valid?

It says that a polynomials can only be constructed with addition, divison and substraction. The square rooth of two is a number, like any other, which can be used to construct a polynomial with these basic operation. Why can't we consider it in the same manner with square rooths of variables ????

Thank you.

## #2 2013-08-21 09:19:32

bobbym

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### Re: Question on the square rooth of polynomials

Each term consists of the product of a constant -- called the coefficient of the term and a finite number of variables (usually represented by letters), also called indeterminates, raised to whole number powers.

The above is a definition. We agree to abide by it to play the game called mathematics. It is a rule.It was a wise decision to define them like that.

(√a + √b) is not a polynomial because the variables a and b are not raised to a positive integer value.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #3 2013-08-21 09:29:53

Al-Allo
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### Re: Question on the square rooth of polynomials

Mmmh... Okay. I just find it weird that we can't take the square rooth even if it doesn't have a number (two in this case) raised as a power...

## #4 2013-08-21 09:42:59

bobbym

Online

### Re: Question on the square rooth of polynomials

We could take that square root but it just would not be a polynomial.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #5 2013-08-21 10:33:55

Al-Allo
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### Re: Question on the square rooth of polynomials

#### bobbym wrote:

We could take that square root but it just would not be a polynomial.

Yes, but that's what I find weird... I don't see the logic behind it. But if it's the way it is, to play the game of mathematics, then we have to consider it that way.

## #6 2013-08-21 14:47:06

bobbym

Online

### Re: Question on the square rooth of polynomials

The decision proved to be a good one. If we allowed negative or fractional powers we would just have another class of functions that are non linear and one we would know very little about.

By using only positive integer powers they now have a class of functions that we know everything about. This is very useful in higher mathematics.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.