(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove by contradiction. Your proof should be based only on properties of the integers, simple algebra, and the definition of rational and irrational.

If a and b are rational numbers, b does not equal 0, and r is an irrational number, then a+br is irrational.

2. Relevant equations

rational numbers are equal to the ratio of two other numbers

3. The attempt at a solution

I wrote a proof but am not sure it is correct. Please tell me what I did wrong and show me the way to do it right if this is not correct. My teacher indicated that we need to make use of the fact that b does not equal 0 (from a+br). Did I do that sufficiently as well?:

Proof:Suppose not. That is suppose that there exists rational numbers a and b, b does not equal zero, and irrational number r such that a+br is rational [We must deduce a contradiction].

By definition of rational, a = c/d, b= e/f , a+br = g/h for some integers c,d,e,f,g,and h with h,f,d, and b not equal to 0.

By substitution, a+b(r) = c/d +(r)( e/f) = g/h.

Solving for r gives: r = (fgd-chf) / (ehd)

Now fgd and chf are integers (being products of integers) and ehd does not equal 0 (by zero product property). Thus by definition of rational, r is rational which contradicts the supposition that r is irrational [ Hence the supposition is false and the statement is true].

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# Discrete Math irrational and rational numbers proof

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