# Functions : Proof by Induction

Let n be a natural number, and let f(x) = x^n for all x are members of R.

1) if n is even, then f is strictly increasing, hence one-to-one, on [0,infinity) and f([0,infinity)) = [0,infinity).

2) if n is odd, then f is strictly increasing, hence one-to-one, on R and f(R) = R.

Prove that f is strictly increasing by induction.

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#### Solution Preview

1.)

We have to prove:

f(x) - f(x-1) > 0 for n = even natural number = 2k and x=[0,infinity)

Step 1:

x = 1

f(1) - f(0) = f(1) - f(0) = 1^(2k) - 0^(2k) = 1 ok.

step 2:

Let for x >= 1

f(x) - f(x-1) = x^(2k) - (x-1)^(2k) = m > 0 .....(1)

step 3:

to prove,

f(x+1) - f(x) > 0

because,

f(x+1) - f(x) = f(x+1) - m - f(x-1) (from eqn. 1)

=> f(x+1) - f(x) = (x+1)^(2k) - m - (x-1)^(2k)

=> f(x+1)-f(x) =x^(2k) +C(2k,1).x^(2k-1) + ....+C(2k,2k) -m-(x-1)^(2k)

here, C(2k,r) = ...

#### Solution Summary

A function is proven by induction to not be strictly increasing.