Question 131519
I'm assuming that your 'axb' really means 'a times b'


So:

{{{a = b +2}}} and {{{a+b=ab}}}


Substitute into the second equation:


{{{(b+2)+b=(b+2)b}}}


{{{2b+2=b^2+2b}}}


Add -2b to both sides:
{{{2=b^2}}}


So: {{{b=sqrt(2)}}} or {{{b=-sqrt(2)}}}, therefore {{{a=2+sqrt(2)}}} or {{{a=2-sqrt(2)}}}


<i>1. Assume that {{{sqrt(2)}}} is a rational number, meaning that there exists an integer {{{a}}} and an integer {{{b}}} such that {{{a / b = sqrt(2)}}}. 
2. Then {{{sqrt(2)}}} can be written as an irreducible fraction {{{a / b}}} such that {{{a}}} and {{{b}}} are coprime integers and {{{(a / b)^2 = 2}}}. 
3. It follows that {{{a^2 / b^2 = 2}}} and {{{a^2 = 2 b^2}}}. ({{{(a / b)^n = a^n / b^n}}}) 
4. Therefore {{{a^2}}} is even because it is equal to {{{2 b^2}}}. {{{2 b^2}}} is necessarily even because it's divisible by 2—that is, {{{(2 b^2)/2 = b^2}}} — 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^2 = (2k)^2}}} is equivalent to {{{2b^2 = 4k^2}}} is equivalent to {{{b^2 = 2k^2}}}. 
8. Because {{{2k^2}}} is divisible by two and therefore even, and because {{{2k^2 = b^2}}}, it follows that {{{b^2}}} 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(2)}}} is a rational number must be false. The opposite is proven: {{{sqrt(2)}}} is irrational.</i>  (courtesy Wikipedia)


Since all integers can be represented as the quotient of integers, all integers are rational.  Since {{{sqrt(2)}}} has been proven to be irrational, {{{sqrt(2)}}} cannot be an integer.  Since all integers can be represented as the sum of 2 and some other integer and {{{sqrt(2)}}} has been shown to be other than an integer, {{{2+sqrt(2)}}} and {{{2-sqrt(2)}}} are also not integers.