Algebra...?

Question

I need help figuring this out...

Prove that there are no non-zero integers a, b such that a^2 = 2b^2. Use the Fundamental Theorem of Arithmetic.

Thanks to anyone that can help me out a little...

Answer 1

The Fundamental Theorem of Arithmetic states that every natural number has a unique prime number factorization.

Proof by contradiction: Suppose there are two numbers a, b such that a²=2b². We have

a = 2^m * p where p is an odd number, and m and p are unique.
b = 2^n * q where q is an odd number, and n and q are unique.

Then
a² = 2^(2m) * p², and 2b² = 2^(2n+1) * q², where p²=q² and
2^(2m) = 2^(2n+1)
→ 2m = 2n+1, an even number being equal to an odd number. Contradiction. Therefore no such pair a, b exists.

Answer 2

The Fundamental Theorem of Alegebra says that there's one and only one way to write out an integer as a product of prime numbers. For example, (3, 3, 11) is the only set of prime numbers which when all multiplied together will give you 99. We're told to use this for the proof, not something else like "the irrationality of the square root of 2".

We can do this proof by "Proof of Contradiction". Assume that there ARE two non-zero integers a and b such that a^2 = 2b^2. Let a = (p1)^(m1) * (p2)^(m2) * (p3)^(m3) *...* (pn)^(mn), where p1, p2, ... pn are the unique prime factors of "a", and m1, m2, etc. are the powers each of them are raised to. Also assume they're indexed such that p1 < p2 < p3 etc. So for the "99" example, 99 = (3^2) * (11^1). Similarly, let b = (q1)^(s1) * (q2)^(s2) * ... * (qn)^(sn).

Since a^2 is 2 times some number, and since 2 is itself a prime number, then p1 = 2. We know a^2 is just "a" times itself, so a^2 = (p1)^(2m1) * (p2)^(2m2) * (p3)^(2m3) *...* (pn)^(2mn). If p1 =2, then a^2=2b^2 becomes

2 * [(p2)^(2m2) * (p3)^(2m3) *...* (pn)^(2mn)] = b^2

None of the prime factors on the left can equal 2, because we said p1, p2, p3 etc. were unique primes. So on the left side we have "2 times an odd number". If b^2 is "2 times some number", then following the same logic as before, the smallest prime factor of b must be 2. So q1 = 2, and dividing both sides by 2 gives you:

(p2)^(2m2) * (p3)^(2m3) *...* (pn)^(2mn) = 2 * [q2^s2 * q3^s3 ... ]

This means "an odd number is 2 times some other number", which can never be true. Therefore, there are no two non-zero integers a and b such that a^2 = 2b^2.

Answer 3

The way I see it, you know that the side with a^2 is a perfect square number. The other side is a perfect square times 2, but it also has to equal a perfect square so that the statement is true. The only way you can get a square number by multiplying times 2 is if it is 2x2 because that is the same as 2^2. Anything else times two doesn't give you a square from a perfect square. So, b^2 would have to equal two. Which means that b would have to be the square root of two to at least get it equal to a square so that it equals to the other side. I hope you understand what I wrote and that it proves it well enough.

Answer 4

The fundamental theorem of arithmetic says that every integer (besides zero, one, and negative one) has a unique factorization into primes.

Let:
a = (p1)^(r1)*(p2)^(r2)*...*(pn)^(rn)
b = (q1)^(s1)*(q2)^(s2)*...*(qm)^(sm)

Where p1,p2,...,pn and q1,q2,...,qm are the prime factors of a and b. When you square a and b, you in effect make all of the powers of these factors double. However, since a^2 = 2b^2, this means if you take the prime factorization of a^2 and multiply it by two, then you get b^2. This is our contradiction. If the prime factors of a^2 are raised to even powers, and we multiply it by 2 (a prime number), the power of 2 in the factorization of the product will be odd. This contradicts that the factorization of b^2 has all even powers.

Answer 5

solve for either a or b
a=(2b^2)^(1/2)

in order for a to be a whole number, 2*b^2 must be a perfect square, but in order for this to happen, b cannot be a whole number.

Answer 6

a^2 = 2b^2
a^2/b^2 = 2
a/b = sqrt(2)
But sqrt(2) is irrational
Therefore a/b cannot be expressed as a rational number.
Hence no values of a and b will satisfy a^2 = 2b^2.