Subject: proof of inequality by mathematical induction
Name: Carol
Who are you: Student

S(n) = 2^n > 10n+7 and n>=10
Basis step is true: S(10) is true
I assume S(k) is true: S(k) = 2^k > 10k+7
Have to show s(k+1) = 2^k+1 > 10(k+1) + 7 is true. This is where I get stuck.


Hi Carol.

Good work so far.

If S(k) > 10k+7, as you've assumed, then

2 x S(k) > 2(10k + 7)

but since
2(10k + 7) = 10k + 10 + 7 + 10k - 3 = 10(k+1) + 7 + 10k - 3

then
2 x S(k) > 10(k+1) + 7 + 10k - 3

and since k ≥ 10, then 10k - 3 > 0, so

10(k+1) + 7 + 10k - 3 > 10(k+1) + 7

So if a > b > c then a > c:

2 x S(k) > 10(k+1) + 7

and since S(k+1) = 2^(k+1) = 2 x 2k = 2 x S(k), we have proven:

S(k+1) > 10(k+1) + 7.

Stephen La Rocque.>