Rational numbers, supremum (Is my proof correct?)

• Incand
I appreciate it.In summary, the set {x ∊ Q; x^2 < 2} has no least upper bound in Q. This can be proven by showing that for every upper bound r1 with r1^2 > 2, there is always a lower upper bound r2 such that r2 < r1. The proof involves setting r2 = r1 - (r1^2 - 2)/(r1 + 2) and showing that r2^2 > 2. This relationship can be used to find a value for k in order to solve for r2. This result shows that the set has no least upper bound in Q, as there is no r ∈ Q that satisfies r^2
Incand

Homework Statement

Show that the set ##\{x \in \mathbf Q; x^2< 2 \}## has no least upper bound in ##\mathbf Q##; using that if ##r## were one then ##r^2=2##. Do this assuming that the real field haven't been constructed.

N/A

The Attempt at a Solution

Attempt at proof:
##r\in Q## is an upper bound of the set if and only if ##r^2 \ge 2##.
Choose any ##r_1 \in \mathbf Q## that satisfies ##r_1^2 >2##. Then ##r_1## is an upper bound of the set.
Set ##r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}>2##.
Note that ##r_2 \in \mathbf Q## and ##r_2 < r_1##. ##r_2## is also an upper bound since ##r_2^2>2##.

This shows that for every upper bound ##r_1## with ##r_1^2 > 2## there's always possible to find an upper bound ##r_2## where ##r_2<r_1##. Hence the only possible least upper bound satisfy ##r^2=2## but there is no ##r\in \mathbf Q## that satisfy that. Hence the set have no least upper bound in ##\mathbf Q##.

Is the above correct? Anything I can do to improve it? I'm also wondering about the choice of ##r_2##. I remembered that from an earlier example in our book but I'm not sure I could've come up with a choice like that by myself without a lot of trial and error. How would I go about finding a choice like that in the first place?

How did you get the following?
Incand said:
##r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}##
When I square ##r_2## I get 2 plus a fraction with the same denominator as yours but a messy quartic in ##r_1## in the numerator.

Incand
Given an upper bound ##r## such that ##r{}^2=2+h## for ##h>0##, you can write an equation for a lower upper bound ##s=r-k## such that ##s^2## is 2 plus some positive fraction of ##h##. Then, using the relationships you've established, solve for ##k##.

andrewkirk said:
How did you get the following?
When I square ##r_2## I get 2 plus a fraction with the same denominator as yours but a messy quartic in ##r_1## in the numerator.
Sorry, I should have clarified that step
##r_2 = r_1-\frac{r_1^2-2}{r_1+2} = \frac{r_1^2+2r_1-r_1^2+2}{r_1+2}=\frac{2r_1+2}{r_1+2}##
square
##r_2^2 = \frac{4(r_1^2+2r_1+1)}{(r_1+2)^2} = \frac{2(r_1^2+4r_1+4)+2(r_1^2-2)}{(r_1+2)^2}= 2+\frac{2(r_1^2-2)}{(r_1+2)^2}##.

andrewkirk said:
Given an upper bound ##r## such that ##r{}^2=2+h## for ##h>0##, you can write an equation for a lower upper bound ##s=r-k## such that ##s^2## is 2 plus some positive fraction of ##h##. Then, using the relationships you've established, solve for ##k##.
Thanks for responding! Is there something missing in my proof where I need this to complete it? When you say the relationship do you mean my value of ##r_2## or something else? I'm not entirely sure I understand.

Incand said:
Is there something missing in my proof where I need this to complete it?
No. I think with the extra steps you've put in, your proof now looks good.

Incand
andrewkirk said:
No. I think with the extra steps you've put in, your proof now looks good.
Thanks for taking the time looking it over!

1. What are rational numbers?

Rational numbers are numbers that can be expressed as a ratio of two integers. This means that they can be written in the form of a/b, where a and b are both integers and b is not equal to 0.

2. How do you know if a number is rational?

A number is rational if it can be expressed as a fraction of two integers. This can be determined by writing the number in decimal form, and if the decimal terminates or repeats, then it is rational. For example, 0.75 can be written as 3/4 and is therefore rational.

3. What is the supremum of a set of rational numbers?

The supremum of a set of rational numbers is the smallest number that is greater than or equal to all of the numbers in the set. This means that there is no other rational number that is larger than the supremum in the given set.

4. How do you prove that a number is the supremum of a set of rational numbers?

To prove that a number is the supremum of a set of rational numbers, you must show that it is greater than or equal to all of the numbers in the set, and that there is no other rational number that is larger. This can be done by using mathematical techniques such as induction, contradiction, or direct proof.

5. Is my proof for the supremum of a set of rational numbers correct?

To determine if your proof is correct, you must make sure that you have followed all the rules and laws of mathematics, and that your reasoning is logical. You can also ask someone else to review your proof and provide feedback or suggestions for improvements.

• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
637
• Calculus and Beyond Homework Help
Replies
2
Views
906
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
937
• Set Theory, Logic, Probability, Statistics
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K