Discrete math. Prove the following statement by contradiction: For all real numbers r and s, if r is rational and s is irrational, then r + 2s is irrational.
Added by Jeremy G.
Step 1
Since r is rational and s is irrational, we can write r = a/b for some integers a and b (with b ≠0), and s cannot be expressed as a ratio of integers. Now, we have r + 2s is rational, so we can write r + 2s = c/d for some integers c and d (with d ≠0). Then, we Show more…
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