Expert: Paul Klarreich - 9/3/2009

Question
Let a,b E R. Prove that if 0<a<1 and b=1-(1-a)^(1/2), then 0<b<a.

Also prove that if 3<a<5 and b=2+(a-2)^(1/2), then 3<b<a.

Answer
Questioner: Cori
Country: United States
Category: Advanced Math
Private: No
Subject: Proofs
Question: Let a,b E R. Prove that if 0<a<1 and b = 1-(1-a)^(1/2), then 0<b<a.

Also prove that if 3<a<5 and b=2+(a-2)^(1/2), then 3<b<a.
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Preliminary theorem: if  0 < x < 1, then which is bigger:  x or sqrt(x)?

Answer: sqrt(x) > x.

Why? If  x < 1, then sqrt(x) < sqrt(1) = 1

then sqrt(x) * sqrt(x) <  1 * sqrt(x)

then  x < sqrt(x),

or  x - sqrt(x) < 0
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Preliminary note:

If  a is between 0,1, then so is 1-a

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Now to the main proof:

Look at b-a.  Is it positive or negative?

b-a = 1 - sqrt(1-a) - a

b-a = 1-a - sqrt(1-a)

b-a = 1-a - sqrt(1-a)  < 0

So  b-a < 0 or  b < a

So a - b > 0,  a > b

I'll leave the  b > 0 to you.
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Second Preliminary: if  1 < x, then which is bigger:  x or sqrt(x)?

Answer NOW:  x is bigger:

Why? If  x > 1, then sqrt(x) > sqrt(1) = 1

then sqrt(x) * sqrt(x) >  1 * sqrt(x)

then  x > sqrt(x),

or  x - sqrt(x) > 0

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Second Preliminary note:

If  a > 3, then  a-2 > 1
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Now,then:

b = 2 + sqrt(a-2)

b-a = 2 + sqrt(a-2) - a

b-a = sqrt(a-2) - a+2

b-a = sqrt(a-2) - (a-2)

But when  a-2 > 1, sqrt(a-2) < a-2, so sqrt(a-2) - (a-2) < 0

b-a = sqrt(a-2) - (a-2) < 0

and  b < a

I leave the  b > 3 part to you.