- #1
sutupidmath
- 1,630
- 4
Well, there is a problem, i have solved/proved it, but i am not sure whether it is correct.
THe problem is this:
Using unique factorization into primes prove that there are no integers a and b such that [tex]a^2=30b^2[/tex], and thus show that [tex]sqrt{30}[/tex] is irrational.
Proof:using unique factorization of any integer greater than 1 or less than -1, we can factor any such integer into the product of powers of distinct primes, or simply into a product of primes.
[tex]a^2=30b^2=>b^2|a^2=>b|a=>\exists k,a=kb[/tex]
Let:
[tex]a=p_1p_2...p_r; b=q_1q_2...q_s[/tex]
[tex]a^2=30b^2=>30=\left(\frac{b}{a}\right)^2=k^2=>\sqrt{30}=k[/tex]
Now from the unique factorization theorem again:[tex]\sqrt{30}=k=d_1d_2...d_n=>30=d_1^2d_2^2...d_n^2[/tex]
=>
[tex]2*3*5=d_1^2d_2^2...d_n^2=>2|d_1^2d_2^2...d_n^2=>2|d_i^2=>2=d_i[/tex]
but this would contradict the unique factorization theorem, and thus this contradiction shows that such a, and b do not exist.
Is this about correct, or there is another way around it?
THe problem is this:
Using unique factorization into primes prove that there are no integers a and b such that [tex]a^2=30b^2[/tex], and thus show that [tex]sqrt{30}[/tex] is irrational.
Proof:using unique factorization of any integer greater than 1 or less than -1, we can factor any such integer into the product of powers of distinct primes, or simply into a product of primes.
[tex]a^2=30b^2=>b^2|a^2=>b|a=>\exists k,a=kb[/tex]
Let:
[tex]a=p_1p_2...p_r; b=q_1q_2...q_s[/tex]
[tex]a^2=30b^2=>30=\left(\frac{b}{a}\right)^2=k^2=>\sqrt{30}=k[/tex]
Now from the unique factorization theorem again:[tex]\sqrt{30}=k=d_1d_2...d_n=>30=d_1^2d_2^2...d_n^2[/tex]
=>
[tex]2*3*5=d_1^2d_2^2...d_n^2=>2|d_1^2d_2^2...d_n^2=>2|d_i^2=>2=d_i[/tex]
but this would contradict the unique factorization theorem, and thus this contradiction shows that such a, and b do not exist.
Is this about correct, or there is another way around it?