Mathematical induction

Discussion in 'General Off-Topic Chat' started by dark_angel, Jul 22, 2008.

Mathematical induction

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Newcomer
2
Jul 14, 2006
Queensland
hi guys,

can u guys help me with a question on my homework, its really been bugging me for a few hours now
the question is, using mathematical induction show that:
2^n < 3^n for any integer n≥1

2. The TeejAlso known as The Tjalian

Former Staff
3
Jun 27, 2004
England
where n = 1

2^1 = 2
3^1 = 3

where n = 2

2^2 = 4
3^2 = 9

...or am I not doing it right?

3. elenarGBAtemp Regular

Member
2
Aug 10, 2007
(2^n)^(1/n)

Newcomer
2
Jul 14, 2006
Queensland
here's an example if it helps its extremely hard to follow because microsoft word wont paste it properly so i apologise for the messiness

P(n): 1^3+2^3+3^3+ …….+n^3=[n(n+1)/2]^2
assume P(n)is true
therefore need to prove P(n+1) is true by making P(n+1)=P(n)
P(n+1): 1^3+2^3+3^3+ …….+n^3+(n+1)^3=[((n+1)(n+1+1))/2]^2
1^3+2^3+3^3+ …….+n^3+(n+1)^3-(n+1)^3=[((n+1)(n+1+1))/2]^2-(n+1)^3
1^3+2^3+3^3+ …….+n^3=(n+1)^2 [((n+2)^2)/4-(4n+4)/4]
1^3+2^3+3^3+ …….+n^3=(n+1)^2 [(n^2+4n+4)/4-(4n+4)/4]
1^3+2^3+3^3+ …….+n^3=(n+1)^2 [(n^2+4n+4-4n-4)/4]
1^3+2^3+3^3+ …….+n^3=(n+1)^2 [n^2/4]
1^3+2^3+3^3+ …….+n^3=(n^2 (n+1)^2)/4
1^3+2^3+3^3+ …….+n^3=[n(n+1)/2]^2
Therefore for every value of n:
1^3+2^3+3^3+ …….+n^3=[n(n+1)/2]^2

5. BigXGBAtemp Regular

Member
1
Aug 29, 2007
over there
2^n < 3^n for any n equal or larger than 1

2^1 < 3^n => 2^2 < 3^n =>...=> 2^x < 3^x
get the xth root on both sides makes (you can get the xth root for any x except 0)
=> 2

Newcomer
2
Jul 14, 2006
Queensland
its just to show that certain patterns that go on forever will always be correct if you keep going down further
instead of becoming incorrect as it goes along

7. Joey90Not around any more

Member
2
Apr 21, 2007
UK
Induction eh?

Well...

when n=1, 2 < 3 so true for n=1

if we assume it is true for n, take the next case:

(2^n)*2 =? (3^n)*3

2^(n+1) =? 3^(n+1)

and because it's true for n, its true for n=n+1

2^(n+1) < 3^(n+1)

now you can induct yourself all the way to infinity.

P.S. It's a kinda retarded thing to do by induction, but if that's what it says...

8. kazumi213GBAtemp Regular

Member
1
Oct 16, 2006
2^n < 3^n for any integer n≥1

log(2^n) < log(3^n)

n * log(2) < n * log(3)

log(2) < log(3)

which is true.

9. PizzaPastaSon of p1ngpong

Member
1
Jan 2, 2008
Akron, Ohio

All of you are communicating in bee language. I'm appalled!

Member
3
Sep 20, 2004
pwned ^^

11. CockroachManScribbling around GBATemp's kitchen.

Member
2
Jan 14, 2006
Brazil
( 2^n < 3^n )

for n = 1: 2 < 3 OK!
for n = 2: 4 < 9 OK!
...
for n = k: 2^k < 3^k

12. deathfisaroNarcistic Deathfisaro Fan

Member
2
Mar 16, 2007
Vancouver, BC
Base case: n=1
2^1 < 3^1, we confirmed that 2^n < 3^n holds for the base case n=1.

Induction:
If 2^n < 3^n, then 2^(n+1) < 3^(n+1), for positive integer n

2^(n+1) = 2^n * 2, 3^(n+1) = 3^n * 3
2^n < 3^n, therefore 2^n * 2 < 3^n * 2

3^n * 3 can be re-written as 3^n * 2 + 3^n.
We know 3^n > 0, and 2^n * 2 < 3^n * 2.
So 2^n * 2 < 3^n * 2 + 3^n still holds true.

Because we proved that 2^n < 3^n with our base case n=1.
We also proved If 2^n < 3^n, then 2^(n+1) < 3^(n+1).
Therefore, 2^n < 3^n is true for any integer n >= 1.

You need to have both the base case and induction.
The base case determines the lower limit of your answer (if you picked n=2 as your base case, your answer becomes 2^n < 3^n is true for any integer n >= 2, even though it's very easy to prove that it is true for n=1).
The induction part has the form of "If (equation involving n) then (equation involving n+1)". "If" is crucial for the induction step, and proving "(equation involving n)" is true with the base case is also crucial because "If false then true/false" is also true and we don't want that.