Comparing two real nubmers

I have been struggling for 2 days now trying to solve these Math problems. I hope someone could solve these in the next few minutes or so.
Subject: Order on R
1) a and b being two non zero numbers and x>0, compare:
a/b and (a+x)/(b+x)
2) a and b being two strictly positive numbers, compare:
(a+b)/ 2 and radical(ab)

Note by Oussama Jaber
4 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

\(2\) is simple. For positive numbers \(a\) and \(b\), \(\frac{a+b}{2}\geq \sqrt{ab}\). This is the two variable case of the Arithmetic Mean-Geometric Mean inequality. Here's a simple proof of the two variable case:

We know that \((a-b)^2\geq 0\)

\(\Rightarrow (a+b)^2 -4ab\geq 0\)

And after a little bit of work, \(\frac{a+b}{2}\geq \sqrt{ab}\).

For \(1\), each one of the following can be true:

\(\frac{a}{b}=\frac{a+x}{b+x}\) [happens when \(x=0\) or \(a=b\)]

\(\frac{a}{b}<\frac{a+x}{b+x}\) [happens when \(x(a-b)<0\)]

\(\frac{a}{b}>\frac{a+x}{b+x}\) [happens when \(x(a-b)>0\)]

Mursalin Habib - 4 years, 5 months ago

Log in to reply

sorry I missed one condition from problem 1.... x>0

Oussama Jaber - 4 years, 5 months ago

Log in to reply

Thank you for your reply, We did not take about the Arithmetic Mean- Geometric Mean inequality. The only method we took when comparing two numbers are:
Compare their difference, compare their squares, compare their radicals ...ect
I managed to to solve 2 by comparing their difference. Here is how I solved it.................................................... a and b are strictly positive, therefore a >0 and b>0

\(\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt{a}-\sqrt{b})^2}{2}\geq0\)
therefore, \(\frac{a+b}{2}\geq\sqrt{ab}\)

But for the problem 1, I knew that there is more than one answer because I plugged in numbers to check for comparison and get different comparison each time I plug in a different number.
But how can use the methods we learned to prove these different comparisons?

Oussama Jaber - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...