# If a and b are real numbers and a≠0, then there exists a unique real number r such that ar + b = 0

Two parts to this question: a) Use a CONSTRUCTIVE proof to show the existence of r and b)Use a proof by CONTRADICTION to show the uniqueness of r

For a) I wrote:

Consider ar+b =0, a≠0 and the premise that the real number r = -(b/a) will cause the equation ar + b to equal 0 for each individual combination of a and b. Now consider a = 2, b=3, and r = -(3/2) Then ar + b = 2(-3/2)+3=0.

For b) I wrote:

Given ar+b=0 and a≠0 suppose ar1 + b = 0 and ar2+b=0 such that r is any real number. These equations can be rearranged so that r1 = -(b/a) and r2 = -(b/a). Thus r1 = -(b/a) = -(b/a) = r2 thus r1=r2. This proves there is a unique r given a real number a and real number b since we assumed two different r yet found they are equal.

Is my thinking correct? This is my first time learning proofs for discrete math so I'm not entirely sure what is considered legal/extensive enough to be correct. My proof for part b seems too straightforward.

• I think what they're expecting you to note is $$ar_1+b=ar_2+b\implies a(r_1-r_2)=0\stackrel{\because a\ne0}{\implies}r_1-r_2=0\implies r_1=r_2.$$
– J.G.
Sep 23, 2019 at 20:33
• For a) I would explicitly substitute $r = b/a$ into the equation; the example may be helpful for you but doesn't form part of the proof. Sep 23, 2019 at 20:34