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Chapter 2: Problem 14

Modify the argument in Theorem \(2.4 .7\) to show that there exists a positive real number \(y\) such that \(y^{2}=3\)

Short Answer

Expert verified
The positive real number \(y\) such that \(y^2 = 3\) is about 1.73205 (also known as the square root of 3) which can be found by continuous iteration of averaging.

Step by step solution

01

Establish Base Case

Start with a base case. We know that \(1^2 = 1\), which is less than 3, and \((2)^2 = 4\), which is greater than 3. Thus, the number we are looking for, \(y\), falls between 1 and 2.
02

Use Average

Using the average of these two numbers will get us closer to \(y\). Find the average \(1.5 = \frac{1+2}{2}\), and square the average to get \(2.25 = 1.5^2\), which is still less than 3.
03

Narrow the Range

Determine that our number is between 1.5 and 2. Take the average of the lower and upper bound \(1.75 = \frac{1.5+2}{2}\), and square it to get \(3.0625 = 1.75^2\), which is more than 3. Now we know that our number is between 1.5 and 1.75.
04

Continue Until Desired Accuracy

Following this process, we get closer and closer to the exact value of \(y\) with each iteration. If this process is repeated infinitely many times, we will arrive at a value of \(y\) which, when squared, equals exactly 3 due to the properties of our real number system.

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Most popular questions from this chapter

(a) Show that if \(x, y\) are rational numbers, then \(x+y\) and \(x y\) are rational numbers. (b) Prove that if \(x\) is a rational number and \(y\) is an irrational number, then \(x+y\) is an irrational number. If, in addition, \(x \neq 0\), then show that \(x y\) is an irrational number.

If \(y>0\), show that there exists \(n \in \mathbb{N}\) such that \(1 / 2^{n}

If \(0

Let \(S\) be a nonempty bounded set in \(\mathbb{R}\). (a) Let \(a>0\), and let \(a S:=\\{\) as: \(s \in S\\}\). Prove that $$ \inf (a S)=a \inf S, \quad \sup (a S)=a \sup S $$ (b) Let \(b<0\) and let \(b S=\\{b s: s \in S\\}\). Prove that $$ \inf (b S)=b \sup S, \quad \sup (b S)=b \inf S $$

Let \(S_{3}=\\{1 / n: n \in \mathbb{N}\\} .\) Show that sup \(S_{3}=1\) and inf \(S_{3} \geq 0 .\) (It will follow from the Archimedean Property in Section \(2.4\) that inf \(\left.S_{3}=0 .\right)\)
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