Can you prove the following statements?

Question

(i) For all real numbers y, b, m with m≠0, there is a unique real number x such that y=mx+b

(ii) For all real numbers y, m, there exist b, xЄR such that y=mx+b

Answer 1

(i) If y=mx+b then
→ mx = y-b
→ x = (y-b)/m which is unique given y, m, b. Then

mx+b = m(y-b)/m + b
= (y-b) + b
= y.

(ii) if m≠0, pick any b and the above proof applies.
If m=0, then b=y will do.

Answer 2

(i) Assume y, b, and m are real numbers, and assume m is not equal to 0.
{We want to prove that there is a unique real number x such that y = mx + b}

Let's try and prove this by contradiction. Assume there is NOT a unique number x such that y = mx + b. Then, it follows that there exists at least two real numbers x1 and x2 such that

y = m(x1) + b, and
y = m(x2) + b

Subtracting both of these equations,

y - y = [m(x1) + b] - [m(x2) + b]

Simplifying this,

0 = m(x1) + b - m(x2) - b

0 = m(x1) - m(x2)

Now, let's factor out m.

0 = m(x1 - x2)

Now that we have two factors equated to 0, it follows that

m = 0, or
x1 - x2 = 0

However, m = 0 cannot be true, since m is specified as non-zero.

Also, if x1 - x2 = 0, then x1 = x2, which also cannot be true, because x1 and x2 are distinct real numbers.

This is a contradiction. Therefore, there has to be a unique real number x such that y = mx + b.

Answer 3

y=mx+b is the equation of a line. If m<>0 then the line isn't horizontal. Every line on the x,y graph through a given y intersept b has a different x for each y.

Follow similar reasoning for (ii).