can you solve this? (1 Viewer)

[Lazarus]

Originally posted by ~*HSC 4 life*~
is general solutions a 3u topic?

Yes.

 

[tooheyz]

hmmm, just another questions guys (sorry for being ever so annoying!)

prove, by induction, that for all positive integers n

2^n >= n + 1

(two to the power of n is greater than or equal to (n+1)

its 3u, but umm yeah, i didnt want to make another thread...

 

[kimmeh]

let n = 1
LHS : 2^1 = 2
RHS : 1 + 1 = 2
.: true for n = 1

assume true for n = k
2^k >= k + 2

RTP n = k + 1 is true

now 2^(k+1) >= k + 2
2^k x 2^1 >= k + 2
(k+1) x 2 >= k + 2 (from assumption)
2k + 2 >= k + 2

.: true

i hope this is right. you kinda jerked my mind again because i havent touched this for a while.. and i need to as well

 

[tooheyz]

Originally posted by kimmeh

2k + 2 >= k + 2

.: true

yeah thats exactly what i got, but i was told my someone that the answers HAD TO EQUAL

i just wanted to check
but yeah it says greater than or equal to...

hmm thanks mate

 

[kimmeh]

it doesnt does it ?
as long as the statement is ture, then you have proved it..

//edit: tooheyz, how come your doing maths again anyway ? arent you due for uni soon ?

 

[tooheyz]

yeah man bridging course

 

[Xayma]

Originally posted by kimmeh
it doesnt does it ?

Nope inequalities dont have to equal which means you will do things like drop numbers from either side eg 7>=4 and 5>=4 etc just so you get k+1

Originally posted by kimmeh
assume true for n = k
2^k >= k + 2

Umm where did you get the +2 from isnt it +1

 

[tooheyz]

Originally posted by Xayma

Umm where did you get the +2 from isnt it +1

2^n >= n+ 1

so for n = k+1

RHS = (k+1) + 1 = k+2

 

[KeypadSDM]

Originally posted by Xayma
Umm where did you get the +2 from isnt it +1

It's a typo. I didn't want to point it out because I thought it was going to confuse them. Well, it did.

 

[tooheyz]

Originally posted by KeypadSDM
It's a typo. I didn't want to point it out because I thought it was going to confuse them. Well, it did.

alright, im confused.

 

[Xayma]

Originally posted by tooheyz
alright, im confused.

It appears when n=k, it it should be 2^k>=k+1
since you're directly substituting n for k. It doesnt change the result but still

 

[Calculon]

How come you double posted instead of just editing?

 

[kimmeh]

Originally posted by tooheyz
yeah man bridging course

i see they try to teach you the whole ext 1 course in 2 weeks? do they give you a test after ?

Originally posted by tooheyz
alright, im confused.

so am i !

 

[Xayma]

Originally posted by Calculon
How come you double posted instead of just editing?

I thought I edited, so I deleted the double post

 

[KeypadSDM]

Originally posted by kimmeh
let n = 1
LHS : 2^1 = 2
RHS : 1 + 1 = 2
.: true for n = 1

assume true for n = k
2^k >= k + 2

RTP n = k + 1 is true

now 2^(k+1) >= k + 2
2^k x 2^1 >= k + 2
(k+1) x 2 >= k + 2 (from assumption)
2k + 2 >= k + 2

.: true

i hope this is right. you kinda jerked my mind again because i havent touched this for a while.. and i need to as well

This post should read:

let n = 1
LHS : 2^1 = 2
RHS : 1 + 1 = 2
.: true for n = 1

assume true for n = k
2^k >= k + <b>1</b>

RTP n = k + 1 is true

now 2^(k+1) >= k + 2
2^k x 2^1 >= k + 2
(k+1) x 2 >= k + 2 (from assumption)
2k + 2 >= k + 2

.: true

i hope this is right. you kinda jerked my mind again because i havent touched this for a while.. and i need to as well

 

[tooheyz]

Originally posted by kimmeh
i see they try to teach you the whole ext 1 course in 2 weeks? do they give you a test after ?
so am i !

yeah i reckon aye. one years worth of work into 12 days


nah there isnt a test afterwards, its been alright so far, just a few stuff have been hard - like binomial theroms and compsite and inverse functions - coz they're new...
havent done them before..