show that if a, b, and c are real numbers and a doesn't equal 0. Then there is a unique solution of the equation ax+b=c

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- Sep 30th 2010, 06:53 AMlm6485unique solution
show that if a, b, and c are real numbers and a doesn't equal 0. Then there is a unique solution of the equation ax+b=c

- Sep 30th 2010, 07:13 AMPlato
Any linear function $\displaystyle y=ax+b-c,~a\ne 0$ is injection (one-to-one).

How many times can a line cross the x-axis? - Sep 30th 2010, 11:13 AMlm6485
I have to demonstrate this using a proof. Yet I am confused on how to get going on it.

- Oct 4th 2010, 12:13 PMlm6485
it says to start with an existence proof and then move to a unique proof.

- Oct 4th 2010, 12:49 PMemakarov
Solve ax+b=c for x. Let's call the solution x0.

Existence claim: There exist an x such that a * x + b = c. Proof: Make sure that x0 is well-defined. Here it means that one does not have to divide by 0 to calculate it. Substitute x0 for x and verify that both sides of the equation are equal.

Uniqueness claim: There is at most one x such that a * x + b = c. Proof: Assume there are two numbers x1 and x2 that satisfy this equation. Substitute x1 for x and then x2 for x; you'll get two equations. Subtract one from the other. You should get a * (x1 - x2) = 0. From here and the fact that a <> 0, conclude that x1 = x2.