Proving 0 < b < a for Given Equations in Intro to Analysis Book

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In summary, the question asks to prove that if 0 < a < 1 and b = 1 - √(1 - a), then 0 < b < a. The hint provided is to use the result 0 < a < 1 implies 0 < a^2 < a. Starting with b = 1 - √(1 - a), we can use the hint to arrive at a contradiction if the proposition 0 < b < a is false. This leads to the conclusion that the actual inequality must hold.
  • #1
iamalexalright
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Homework Statement


This is from Wade's Intro to Analysis book; problem 5, Chapter 1.

Prove that if [tex]0 < a < 1[/tex] and [tex] b = 1 - \sqrt{1 - a} [/tex] then [tex]0 < b < a[/tex]


Homework Equations



The book hints to use this result: [tex]0 < a < 1[/tex] implies [tex]0 < a^{2} < a[/tex]

The Attempt at a Solution


First I substitute b into the equation to get:
[tex]0 < 1 - \sqrt{1 - a} < a < 1[/tex]

By the additive property I add [tex]-1[/tex] and get:
[tex]-1 < -\sqrt{1 - a} < a - 1 < 0[/tex]

Using the multiplicative property I multiply by [tex]-1[/tex] and get:
[tex]0 < 1 - a < \sqrt{1 - a} < 1[/tex]

That above bears resemblance to the hint equation but I don't know exactly where to go (and my inkling of an idea doesn't seem to prove the initial statement).
 
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  • #2
Hi iamalexalright! :smile:

(have a square-root: √ :wink:)
iamalexalright said:
Prove that if [tex]0 < a < 1[/tex] and [tex] b = 1 - \sqrt{1 - a} [/tex] then [tex]0 < b < a[/tex]

First I substitute b into the equation to get:
[tex]0 < 1 - \sqrt{1 - a} < a < 1[/tex]

But that's the answer!

Start with the question: b = 1 - √(1 - a).

Now use the hint. :wink:
 
  • #3
hrm, and you'll have to forgive me but I am new at this :)

[tex]b = 1 - \sqrt{1 - a}[/tex]
[tex]b^{2} = 2 - 2\sqrt{1-a} - a = 2b - a[/tex]

This certainly feels like the wrong direction (or I just can't find the next logical step). If I go from here I have trouble comparing a to b ...

If I could get a bigger nudge in the right direction I'd be much appreciative (but nothing more than a nudge please)
 
  • #4
Alright, let's first suppose that the proposition that [itex]0 < b < a[/itex] is false. Then clearly we must have that [itex]0 < a \leq b = 1 - \sqrt{(1-a)}[/itex]. From this, we know that [itex]0 < \sqrt{(1-a)} < 1 - a[/itex]. Now, can you use the hint to arrive at a contradiction? What does this contradiction suggest about the actual inequality?
 
  • #5
ah cool, I see it now. For some reason I don't like proofs by contradiction (I really can't say that since I'm just starting to write them). For some reason it's tougher for me to wrap my head around them. Anyway, thanks for the help you two!
 
  • #6
(just got up :zzz: …)
iamalexalright said:
… For some reason I don't like proofs by contradiction …

Nor do I! :wink:

But you don't need it here …

just start with 1 - b = √(1 - a), and use the hint. :smile:
 
  • #7
iamalexalright said:
ah cool, I see it now. For some reason I don't like proofs by contradiction (I really can't say that since I'm just starting to write them). For some reason it's tougher for me to wrap my head around them. Anyway, thanks for the help you two!

Yes trying a proof by contradiction for basic inequalities in analysis is ugly, but it can sometimes get the job done well. Once you get to least upper bounds or basic metric topology, you might find contradiction approaches more congenial or elegant.
 

FAQ: Proving 0 < b < a for Given Equations in Intro to Analysis Book

How do you prove 0 < b < a for given equations in Intro to Analysis book?

To prove this statement, we need to use the properties of inequalities and basic algebraic manipulations. First, we start by assuming that a > 0 and b > 0. Then, we can multiply both sides of the inequality a > b by a to get a^2 > ab. Similarly, we can multiply b > 0 by b to get b^2 > 0. Combining these two inequalities, we get a^2 > ab > b^2. Since both a^2 and b^2 are positive numbers, we can take the square root of both sides to get a > b. Therefore, we have proved that 0 < b < a.

Why is it important to prove 0 < b < a for given equations in Intro to Analysis book?

Proving this statement is important because it is a fundamental concept in analysis and is used in many other theorems and proofs. It also helps to establish the relationship between two numbers and their magnitudes, which is crucial in understanding the behavior of functions and inequalities.

Can this statement be proved for any given equations?

Yes, this statement can be proved for any given equations as long as the equations satisfy the conditions a > 0 and b > 0. These conditions ensure that the numbers are positive and the algebraic manipulations used in the proof are valid.

Are there any alternative methods to prove 0 < b < a for given equations?

Yes, there are alternative methods to prove this statement. One method is to use the concept of absolute value and the triangle inequality. Another method is to use the properties of exponents and logarithms to show that a > b is equivalent to a^x > b^x for any real number x.

How does this statement relate to the concept of strict inequality?

This statement is a strict inequality, meaning that the inequality symbol (<) is used instead of the less than or equal to symbol (≤). This is because the statement 0 < b < a implies that b is strictly between 0 and a, and is not equal to either of them. This concept is important in analysis as it helps to differentiate between different types of inequalities and their properties.

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