# Simple question on arithmetic.

1. Aug 11, 2013

### cdux

I've noticed 2 x 0.2 x 0.8 happens to give the same result with 1 - (0.2^2 + 0.8^2).

Can the latter be rearranged to the first? And if yes, can someone direct me to the name of the branch of math that describes this problem? It definitely reminds me of something from high school I didn't pay attention to..

2. Aug 11, 2013

### cdux

Oh wait, I guess for this question to avoid being meaningless 0.2 and 0.8 should get the symbols A and B, otherwise it could just be 'rearranged' to 0.32.

3. Aug 11, 2013

### cdux

Hrm, I managed to rearrange it to 1-(A+B)^2 = 0. So that might mean it's a property that ONLY applies when A + B = 1, or at least when that '1' being C is A + B. Am I right?

4. Aug 11, 2013

### cdux

Or wait, also when (A + B)^2 is C, which is rarer.

5. Aug 11, 2013

### jbriggs444

So rearrange the equation that way...

2ab = 1 - ( a2 + b2 )

Now rearrange to put the constant term on the left and the rest of the terms on the right...

1 = a2 + 2ab + b2

If you manage to get it in this form, the right hand side is recognizable

1 = (a+b)2

So you can conclude that as long as a and b add to 1 then the relationship that you observed will hold good. The actual a and b that you chose are 0.2 and 0.8. Those do add to 1.

EDIT: It appears that we crossed posts. You have already realized this.

Last edited: Aug 11, 2013
6. Aug 11, 2013

### cdux

Is my additional assessment that also (a+b)2 could be 1 correct? (Or I guess = C if 1 = C there)

7. Aug 11, 2013

### Enigman

so, 2ab=1-a^2-b^2
a^2+b^2+2ab=1
(a+b)^2=1
'Property' applies when a+b=-1 or +1
Not much of a mystery
Mr.E

8. Aug 11, 2013

### jbriggs444

Actually twice as likely for C positive and impossible for C negative.

[Although it is sloppy to talk about likelihood rigorously without a probability distribution available]

9. Aug 11, 2013

### cdux

True.

I don't get the first part. Why is it more likely? (even if it's loosely described)

edit: Unless you just mean because it excludes the negatives.

10. Aug 11, 2013

### jbriggs444

Yes, that's all I was getting at. That there are two values of a+b for which for (a+b)2=c
as long as c is strictly positive.