Prove that if 0 < a < 1 and b=1-square root of (1-a), then 0 < b < a?

Question

All i can think of is that if you make a plot of y = x and y = 1 - square root of (1-a) from x = 0 to x = 1 you will see that the graph of y = 1 - square root of (1-x) is always beneath the graph of y = x. However, the proof is to be done without using graphs. How is that possible then?

Answer 1

first of all, 0 implies 0 < 1 - sqrt(1-a) < 1
since 0
1-a < sqrt(1-a)
so 1 - (1-a) > 1 - sqrt(1-a)
a > 1 - sqrt(1-a) = b > 0

Answer 2

ok, so you are given that 0
We know that 1-a < 1 since a is, by definition already less than one, so subtracting a number that is less than one from 1 gives a number that must be less than one. Now, since we know that 1-a < 1, we also know that sqrt(1-a) > 1-a, since if 1-a < 1, then 1-a is some fraction, and the square root of a fraction is greater than the original fraction. (an example, 1/sqrt(2) is greater than 1/2).

So we now know that 0 < 1-a < sqrt(1-a).

Multiply that by -1 and you get that 0 > a-1 > -sqrt(1-a)

Add one to all the terms and you get that:

1 > a > 1-sqrt(1-a), but 1-sqrt(1-a) = b, so:

1 > a > b, so now we know that:

b < a, but a is less than 0 and b = 1-sqrt(1-a) which must be greater than 0, so we can deduce that b > 0.

Therefore, 0 < b < a

Answer 3

Prove that if 0 < a < 1 and b=1-sqrt(1-a), then b > a
****(NOT b < a !!!)****
or to prove 0 < 1 - sqrt(1-a) ,a if 0< a < 1

if 0 < a < 1
then 0 < 1-a < 1
and 0 < sqrt(1 - a) < 1 - a < 1
so
- sqrt(1 - a) > - (1 - a) , or
- sqrt(1 - a) > a - 1
1 - sqrt(1 - a) > 1 + a - 1
or
b = 1 - sqrt(1 - a) > a

Answer 4

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Answer 5

If 0 < a < 1 and b=1-sqrt(1-a), then 0 < b < a
Since b = 1-sqrt(1-a) then b-1 = -sqrt(1-a)
So sqrt(1-a) = 1-b
So a>b since sqrt(1-x)>1-x for 0 Thus 0

Answer 6

You could give random values for a
like:
a= 0.004
a=0.993
a=0.5

I'm not sure if you're allowed to do so though

Answer 7

? sorrry i dont know and im bad at math

Answer 8

x=38 duh