# Domain: deceivingly simple, turns evil

StephenPrivitera
I wonder what went wrong here:
Find the domain of f(x)=sqrt(1-sqrt(1-x2))
So of course the term under the radical must be greater than or equal to zero. For neatness I ignore the equal part.

1-sqrt(1-x2)>0
1>sqrt(1-x2)
both left and right are positive, so
1>1-x2
x2>0

But it seems to me that the domain is actually [-1,1]

You can do this alternatively by finding the domain of the innermost radical, which is [-1,1], and intersecting it with the set of all x such that sqrt(1-x2) is in the domain of h(y)=sqrt(1-y). Since sqrt(1-x^2)>=0 for all x in its domain, the intersection is just [-1,1]. So what happened wrong with the first approach?

## Answers and Replies

Staff Emeritus
Gold Member
Originally posted by StephenPrivitera
1-sqrt(1-x2)>0
1>sqrt(1-x2)
both left and right are positive, so
1>1-x2

You cannot square both sides of an inequality, because that implies that both sides are equal!

See, it works for an equation as follows:

(1-x2)1/2=1
((1-x2)1/2)2=12

Note carefully that what we are really doing is:

(1-x2)1/2*(1-x2)1/2=1*1

But remember our rule that "whatever you do to one side, you have to do to the other side".

Question: How is it that we can square both sides, when we are multiplying the left by (1-x2)1/2 and the right by 1?

Answer: Because in the beginning we asserted that those two quantities are equal, a condidtion we do not have in the inequality case.

StephenPrivitera
I suspected that might have been the case. But then I thought, if we write a>b as a=b+c for some c in R, where a,b,c are greater than 0, then
a2=b2+(2bc+c2)=b2+d. d is positive and in R. So, by my definition, a2>b2.

Staff Emeritus
Gold Member
Originally posted by StephenPrivitera
I suspected that might have been the case. But then I thought, if we write a>b as a=b+c for some c in R, where a,b,c are greater than 0, then
a2=b2+(2bc+c2)=b2+d. d is positive and in R. So, by my definition, a2>b2.

Yeah, but you squared an equation, and then drew an inference about an inequality, which is perfectly valid. At no point did you perform the invalid step of squaring the inequality.

StephenPrivitera
Yes, but doesn't this show that if a>b and a>0 and b>0 then a2>b2?
So, since 1>sqrt(1-x2)
and 1>0 and sqrt(1-x2)>0 for all x in it's domain
then 1>(1-x2).
I suspect that the problem lies in the fact that I have to comment
sqrt(1-x2)>0 for all x in it's domain.
I believe that when I get to x2>0, this means that the inequality holds not for all x but for for all x in the domain of sqrt(1-x2). I must be restricting the values of x when I decide to square each side.
I understand that in general you cannot square an inequality. For example, -5>-10 does not imply 25>100. But as far as I can tell squaring an inequality is ok if both the left and right are positive. Of course, I've been wrong before .

Staff Emeritus
Gold Member
Originally posted by StephenPrivitera
Yes, but doesn't this show that if a>b and a>0 and b>0 then a2>b2?

Ah, you're right, it does.

But something is going screwy, because I can just as easily prove that the domain is [-1,1] if I only use:

1>(1-x2)1/2

Because if I multiply the right by (1-x2) and the right by 2, then I certainly can preserve the inequality:

2>=(1-x2)

and get x2=<1, which leads to [-1,1]. In fact, you can pretty much make the domain anything you want if that inequality is all you use. Just multiply the left by some number a>0, and the right by some number 0<b<a.