Let R be the set of real numbers, Q be the set of rationals and R - Q be the irrationals.
Property of rational numbers Q:
1. Any rational number can be represented in the form of p/q where p and q are relatively prime integers, , and q is not equal to 0
2. Addition or subtraction of two rational numbers give another rational number because Q is an ordered field and it is closed for addition. Similarly it is also closed for multiplication.
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We will use contradiction method to prove that √2 + √3 is an irrational number.
Let us assume that √2 + √3 be a rational number p/q
√2 + √3 = p/q
multiply and divide by (√2 - √3) in LHS
(√2 + √3) (√2 - √3) / (√2 - √3) = p/q
(2 - 3) / (√2 - √3) = p/q
-1/ (√2 - √3) = p/q
(√3 - √2 ) = q/p [Another rational number]
On subtracting
(√2 + √3) - (√3 - √2) = p/q - q/p
2√2 = p/q - q/p
[take p/q - q/p = r/s another rational number]
2√2 = r/s
LHS (2√2) is an irrational number whereas RHS (r/s) is a rational number.
Therefore r/s must be an irrational number, which implies ( p/q - q/p) must be an irrational number which is possible only if p/q is an irrational number. Thus our assumption that p/q is a rational number is wrong.
Hence √2 + √3 is an irrational number.
