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Number theory Difficulty 6.0 National olympiad Prove it

Theorem 10.1. Let α\alpha and β\beta be rational numbers. Then α+β,αβ,αβ\alpha+\beta, \alpha-\beta, \alpha \beta, and α/β\alpha / \beta (when β0\beta \neq 0 ) are rational.

Solution

Proof. Since α\alpha and β\beta are rational, it follows that α=a/b\alpha=a / b and β=c/d\beta=c / d, where a,b,ca, b, c, and dd are integers with b0b \neq 0 and d0d \neq 0. Then, each of the numbers
α+β=a/b+c/d=(ad+bc)/bdαβ=a/bc/d=(adbc)/bdαβ=(a/b)(c/d)=ac/bdα/β=(a/b)/(c/d)=ad/bc(β0)\begin{aligned} \alpha+\beta & =a / b+c / d=(a d+b c) / b d \\ \alpha-\beta & =a / b-c / d=(a d-b c) / b d \\ \alpha \beta & =(a / b) \cdot(c / d)=a c / b d \\ \alpha / \beta & =(a / b) /(c / d)=a d / b c \quad(\beta \neq 0) \end{aligned}
is rational, since it is the quotient of two integers with denominator different from zero.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.