induction with inequalities...

[killerfish]

1. The problem statement, all variables and given/known data

Prove: 2n + 1 < 2n , with n >= 3


2. Relevant equations



3. The attempt at a solution

2 (3) + 1 = 7 and 23 = 8.
So 2 (3) + 1 < 23.
Thus the inequality holds with n = 3:
Suppose the inequality holds with n = k
Then 2k+ 1 < 2k:
So 2k + 1 + 2 < 2k + 2
2k + 3 < 2k + 2k
2k + 3 < 2(2k)
2 (k + 1) + 1 < 2(k+1):
So, the inequality holds with n = k + 1:






Hi guys,

some of the transition on the RHS, im blurred. Like the above there are 2 parts i dont understand,

So 2k + 1 + 2 < 2k + 2
2k + 3 < 2k + 2k

on RHS, how to get from 2k+2 to 2k+2k. Arent when we do a change on LHS(e.g +2), is should be equal to RHS(e.g+2)? sry my understanding for induction is weak, can someone help elaborate the solution...

Thanks very much.

[kuruman]

It's really simple.

You already have that the LHS is less than the RHS

2k + 3 < 2k + 2

If you replace the 2 on the RHS with 2k, a number greater than 2 (k > 1 by assumption), the inequality still holds, therefore

2k + 3 < 2k + 2k.

[HallsofIvy]

2< 2k for any k> 1 and so, adding 2k to both sides, 2k+ 2< 2k+ 2k so 2k+ 2< 2(2k)= 2k+1.