Prove that if the real-valued function f is strictly increasing, then f is oneto-one.

See proof below

Step-by-step explanation:

Let x,y be arbitrary real numbers. We want to prove that if x≠y then f(x)≠f(y) (this is the definition of 1-1).

If x≠y, we can assume, without loss of generality that x<y using the trichotomy law of real numbers (without loss of generality means that the argument in this proof is the same if we assume y<x).

Because f is strictly increasing, x<y implies that f(x)<f(y). Therefore f(x)≠f(y) because of the trichotomy law, and hence f is one-to-one.

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5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10