Question

Prove by induction
2n + 3 ≤ 2^n
for all integers n ≥ N. You should find the minimal integer value
of N for which this proposition is true and then construct the proof using this
value.

Answer 1

please?

Answer 2

probably true if \(n=4\) right ?

Answer 3

well actually since it is \(\leq \) it is true if \(n=3\)

Answer 4

N = 4 not 3

Answer 5

satellite73, can't believe you are still here, you are the one who helped me pass analysis haha!
i haven't done induction in a while so i forgot how to do this especially since they are teaching us to only use one side of the equation in this module. Yes N would be 4

Answer 6

i am glad you passed

Answer 7

assume true for \(k\) that is assume \(2k + 3 ≤ 2^k\) now lets try it for \(n=k+1\) you get
\[2(k+1)+3=2k+5\] and by induction \(2k+3<2^k\) so \(2k+5<2^k+2<2^{k+1}\)

Answer 8

i don't understand the last line

Answer 9

The so bit

Answer 10

Now prove 2^k + 2 <= 2^(k+1)... note n >= 4 so 2^n >= 16

Answer 11

where is the 2^k + 2 coming from

Answer 12

2k + 3 < 2^k therefore 2k + 5 < 2^k + 2; the goal is to prove 2k + 5 < 2^(k+1)

Answer 13

<= ***

Answer 14

so if we know that 2k + 5 < 2^k + 2 is true
and we have to prove that 2k + 5 < 2^(k+1) is true
then do i have to prove that 2^k+2=2^(k+1) am i understanding this correctly? haven't done induction with inequalities before this is a total mess :/

Answer 15

No, you want to prove 2^k + 2 <= 2^(k+1) :-)
If k >= 4, then 2^k >= 16, right? Therefore 2^k + 2 <= 18 <= 32 <= 2^(k+1)

Answer 16

thanks for your help