Question

Struggling to understand how the fundamental theorem of arithmetic should be used to prove irrationality.

Answer 1

I can prove it for root 2 however something like the cube root of 7 still eludes me.

Answer 2

i write cube root 7 = a/b
7=a^3/b^3
b^3*7=a^3

then im slightly stuck. there should be a contradiction but i cannot quite see it

Answer 3

What is the fundamental theorem?

Answer 4

every integer >1 can be expressed as a unique product of primes

Answer 5

Why are you exploring this with cubes and roots?

A prime is evenly divisible by only itself and 1.

1 prime
2 prime
3 prime
4 = 2*2
5 prime
6 = 3*2

Perhaps you can prove this by induction. Or by contradiction:
Could two different products of primes be the same number?

Answer 6

I an trying to prove that the 3rd root of 7 is irrational by contradiction.

Answer 7

\[\sqrt[3]{7}\]?

Answer 8

suppose it is rational
\[\sqrt[3]{7}=\frac{ a }{ b }\rightarrow \\to \ the \ power \ of \ three\\7=\frac{ a^3 }{ b^3 }\]
\[a^3=7b^3 \rightarrow a=7q\\\]
now a=7q
insert in formula
\[(7q)^3=7b^3\\b^3=49q^3=7(7q^3)\\so\\b=7k\]
b must be 7k
so it is contradiction
because a rational number does not simplify
Q={a/b: a,b elongs to Z ,b not zero , (a,b)=1 }
(a,b)=1 means a/b does not simplify
like this
2/4 is not rational
but
1/2 is rational

Answer 9

sorry. I don't see how that used the fundamental theorem of arithmetic (FTA). I suspect a question will come up on my exam about using the FTA to prove irrationality of something but I'm never completly sure about how to do that.