# if root2 is an irrational number,then prove that root2+root3 is also an irrational number.

A rational number can be represented in p/q form where p and q have no common factors and  q ≠ 0 . Let us assume here $\sqrt{2}$$+\sqrt{3}$  is a rational number and thus it can be represented in p/q form so,

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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.

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