Proving Existence of Two Numbers with Difference at Most 10

[rbzima]

Homework Statement

The sum of 5 positive real numbers is 100. Prove that there are two numbers among them whose difference is at most 10.

Homework Equations

Nothing really...

The Attempt at a Solution

The biggest problem I'm running into is that I can think of specific examples, but translating that into an algebraic argument has always been my weak area. Getting started is where I struggle the most... but I'm thinking the following:

Let's assume there are no two positive real numbers whose difference is at most 10.
Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5} each represent some positive real number whose sum equal 100. Given that a_{1} is the smallest real number, a_{2} = 10 + a_{1}, a_{3} = 10 + a_{2} and so on...

Don't really know where to go from here! Suggestions?

 

[quantumdude]

Well you've got a finite set of numbers, so you know it has a minimum. Let x_1 be that minimum. Furthermore they're real numbers so they're ordered. Let x_1\leq x_2 \leq x_3 \leq x_4 \leq x_5. Now suppose that there do not exist a pair of numbers such that their difference is at most 10.

Try to show that x_1+x_2+x_3+x_4+x_5>100.

 

[rbzima]

Well you've got a finite set of numbers, so you know it has a minimum. Let x_1 be that minimum. Furthermore they're real numbers so they're ordered. Let x_1\leq x_2 \leq x_3 \leq x_4 \leq x_5. Now suppose that there do not exist a pair of numbers such that their difference is at most 10.

Try to show that x_1+x_2+x_3+x_4+x_5>100.

Wow, that was straight forward... Totally figured it out...
I had an epiphany right before you responded, so I should be good now. I was definitely thinking the same thing, and ended up solving in terms of a_{1}. Regardless of how small a_{1} is, as long as it's not 0, you will receive some value that is greater than 100, thus we have a contradiction, which means that there must be at least two values who have a difference of at most 10!

Thanks for the help!