Answer: D) An irrational number
rational + irrational = irrational
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Proof:
The claim is that A+B = C where,
- A = some rational number
- B = some irrational number
- C = some other irrational number
Because A is rational, we can write it like A = p/q for some integers p,q. The value of q cannot be zero.
Let's for a moment consider the opposite scenario and let's assume that A+B was rational. We'll prove that a contradiction arises from this (hence this is a proof by contradiction).
If A+B = C was rational, then C = r/s for some integers r,s. The s cannot be zero.
From there we can have these steps to isolate B
A+B = C
B = C - A
B = (r/s) - (p/q)
B = (rq - ps)/(qs)
B = (some integer)/(some other nonzero integer)
B = some rational number
But this directly contradicts B set up as an irrational number.
So if A is rational and B is irrational, then A+B cannot possibly be rational due to the proof by contradiction above. The only other possibility is that A+B must be irrational.