The cube root of 2 is irrational true or false? **please provide proof**
true
it's irrational. i think you need to learn calculus to find the proof there's a rule i know that if a root isn't a perfect root, then it's a perfect square
no u can give proof that decimal representation of it is non terminating and non recurring
it's irrational bccasue it's can't be put in the form of a/b , where a,b are real nembers
1. suppose is rational so it is= p/q where p and q are integers with no common divisor. 2. raise to exponent of 3 and we have 2 = (p/q)^3 or 2q^3=p^3 3. in LHS we have an even number , so in RHS q must be even. let say p= 2 r where r is an integer. 4. Substitute in 2 q^3=p^3 we have 2 q^3=(2r)^3 or 2 q^3=8 (r^3) or if we divide by 2, q^3=4 (r^3). 5. Now in the RHS we have an even number, so the LHS must be even or q = 2 s, where s is an integer. 6. From the last relation (q = 2 s) and p= 2 r (obtained above), we conclude that q and p have 2 as a common divisor. 7. The steps 1 and 6 are contradictory.
ok
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