SOLUTION: a=b+2 a+b=axb show that a and b are not integars

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Question 131519: a=b+2
a+b=axb
show that a and b are not integars
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that your 'axb' really means 'a times b'

So:
a+=+b+%2B2 and a%2Bb=ab

Substitute into the second equation:

%28b%2B2%29%2Bb=%28b%2B2%29b

2b%2B2=b%5E2%2B2b

Add -2b to both sides:
2=b%5E2

So: b=sqrt%282%29 or b=-sqrt%282%29, therefore a=2%2Bsqrt%282%29 or a=2-sqrt%282%29

1. Assume that sqrt%282%29 is a rational number, meaning that there exists an integer a and an integer b such that a+%2F+b+=+sqrt%282%29.
2. Then sqrt%282%29 can be written as an irreducible fraction a+%2F+b such that a and b are coprime integers and %28a+%2F+b%29%5E2+=+2.
3. It follows that a%5E2+%2F+b%5E2+=+2 and a%5E2+=+2+b%5E2. (%28a+%2F+b%29%5En+=+a%5En+%2F+b%5En)
4. Therefore a%5E2 is even because it is equal to 2+b%5E2. 2+b%5E2 is necessarily even because it's divisible by 2—that is, %282+b%5E2%29%2F2+=+b%5E2 — and numbers divisible by two are even by definition.)
5. It follows that a must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
6. Because a is even, there exists an integer k that fulfills: a+=+2k.
7. Substituting 2k from (6) for a in the second equation of (3): 2b%5E2+=+%282k%29%5E2 is equivalent to 2b%5E2+=+4k%5E2 is equivalent to b%5E2+=+2k%5E2.
8. Because 2k%5E2 is divisible by two and therefore even, and because 2k%5E2+=+b%5E2, it follows that b%5E2 is also even which means that b is even.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).
Since there is a contradiction, the assumption (1) that sqrt%282%29 is a rational number must be false. The opposite is proven: sqrt%282%29 is irrational. (courtesy Wikipedia)

Since all integers can be represented as the quotient of integers, all integers are rational. Since sqrt%282%29 has been proven to be irrational, sqrt%282%29 cannot be an integer. Since all integers can be represented as the sum of 2 and some other integer and sqrt%282%29 has been shown to be other than an integer, 2%2Bsqrt%282%29 and 2-sqrt%282%29 are also not integers.