Prove that ^3sqrt(4) is irrational.

It is good getting practice on these Chris.  I did one the other day that was really hard.

Thanks for the question to Mellie.

This is best done as a "proof by contradiction"

Let us suppose that ∛4  can be expressed as the rational  a/b   where a, b are relatively prime integers -  [ have no common factors ]

So we have 

∛4  = a/b        cube both sides

4   = a^3  / b^3       multiply both sides by  b^3

4b^3   =  a^3        since the left side is even, then a^3 must also be even.....and since the product of an odd integer cubed would also be odd, then "a" itself must be even

Then, since  "a" is even,  and we can write any even  number as  "2m,"  let us write "a"  as "2m"    ....and we have

4b^3   = (2m)^3    simplify

4b^3  = 8m^3       divide both sides by  4

b^3  = 2m^3        note that the right side is even, so this would imply that b^3 is also even

And, as before......if b^3 is even, then so is "b"

But if "a"  is even   and "b"  is even, then they both have a common  factor of 2.......but......this violates our assumption that a and b are relatively prime......so....our assumption that ∛4  can be expressed as the rational  a/b  must be incorrect..... and .this means ∛4  is irrational........