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.
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)^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?