Question

Prove by induction that if \[0 < a < 1\]then \[(1-a)^n \geq 1-na\]

Answer 1

I think you know how to do it

Answer 2

Let P(n) be the statement \[(1-a)^n \geq 1 - na; \ \ \ 0 < a < 1\]

Answer 3

P(1) is the statement \[(1-a)^1 \geq 1 - (1)a\]iff\[1-a \geq 1-a\]which is true

Answer 4

Now prove that given a Pk true... Pk+1 is true as well! :)

Answer 5

If all of P(1),..., P(k), are true, then in particular, P(k) is true. It must be shown that P(k+1) is true.
P(k) is the statement \[(1-a)^k = 1-ka ; \ \ \ 0 < a < 1\]

Answer 6

oops it's the staement \[(1-a)^k \geq 1 - ka ; \ \ \ 0 < a < 1\]

Answer 7

let's multiply both sites by 1-a

Answer 8

\[(1-a)^k \geq 1 - ka\]iff \[(1-a)^k(1-a)\geq(1-ka)(1-a)\]iff\[(1-a)^{k+1}\geq1 -a -ka+ka^2\]iff\[(1-a)^{k+1}\geq\]im stuck :(

Answer 9

ka^2 is an always positive term, how can that help you out?

Answer 10

hmm.. it can be shown that (1-a)^{k+1} is bigger than the right hand side?

Answer 11

You agree that if you have...
a > b+something positive => a > b ?

Answer 12

yea
\[(1-a)^{k+1}\geq1 -(k+1)a+ka^2\]

Answer 13

so how do I end the proof

Answer 14

since k is in N, and a^2 > 0, then ka^2 > 0, etc. etc.

Answer 15

so I simply say
\[(1-a)^{k+1}\geq1 -(k+1)a+ka^2\]iff\[(1-a)^{k+1}\geq1 -(k+1)a\]is that alright?

Answer 16

You wanted to prove that...
(1-a)^k >= 1-ka.
Now, since ka^2 is always positive, you can just take it... yeah, you got it ;)

Answer 17

... which proves that P(k+1) is true.

Answer 18

right!