# Discrete Mathematics Proof Problem

1. Sep 9, 2010

### tuttleforty77

1. The problem statement, all variables and given/known data
Which is larger, square root of 2 or cubed root of 3? Prove one is larger than the other without using decimal approximations for either number.

3. The attempt at a solution
I attempted to solve this through the contradiction that they were even. If they are not even then obviously one has to be larger than the other.

Assume sqrt(2) = cuberoot(3), which is the contradiction to our original problem. We can manipulate this expression using algebra to evaluate whether it is true or not.

sqrt(2)^3 = cuberoot(3)^3
2*sqrt(2) = 3
sqrt(2)= 3/2
sqrt(2)^2 = (3/2)^2
2 = 9/4 This is our false statement!

Since the two numbers are not equivalent, one must therefore be larger than the other number.

Now I know they're not equal so one must be larger. Can I conclude from my algebra at the end that the cube root of 3 is larger?

2. Sep 9, 2010

### jgens

It might be better to start with a statement like: Clearly 3/2 > 21/2. And then explicitly show that 31/3 > 21/2.

3. Sep 9, 2010

### Petek

My approach would be to raise both numbers to the same power such that the results are both integers. It's then easy to see which integer is larger and hence which of the original numbers is larger.

Petek

4. Sep 9, 2010

### tuttleforty77

That's a good point, I can skip all the messy algebra if I just raise them to the 6th power. That would result in 3^2 = 9 and 2^3 = 8. Seems too easy but that definitely works