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The homework question is "Prove that if and are rational numbers then 2 + 3 is a rational number"

So I wrote this to try to prove it:

*Let r and s be rational numbers

By definition of rational, r = a/b and s = c/d and a, b, c, and d are integers with b not equal to 0 and d not equal to 0

Thus, 2r + 3s = 2a/b + 3c/d by substitution

2a/b + 3c/d = (2a)d + b(3c) / bd by basic algebra

Let p = (2a)d + b(3c) and let q = bd. p and q are integers because integers are closed under multiplication and addition, and a, b, c, and d were already declared as non-zero integers

Therefore, 2r + 3s = p/q, and p and q are non-zero integers

Therefore, 2r + 3s is a rational number by the definition of rational.*

I'd appreciate if you could point out any technical errors I made.

davidib17
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1 Answers1

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There are only two small errors that I can see.

  • You write $(2a)d + b(3c) / bd$ when you mean $((2a)d + b(3c)) / bd$.
  • You say that $a, b, c,$ and $d$ were declared as non-zero integers; in fact, only $b$ and $d$ are guaranteed to be non-zero here.

Other than that, your proof is fine.

TonyK
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