Mathematical induction

dark_angel · Jul 22, 2008

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

The Teej · Jul 22, 2008

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?

elenar · Jul 22, 2008

(2^n)^(1/n)

dark_angel · Jul 22, 2008

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

BigX · Jul 22, 2008

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

dark_angel · Jul 22, 2008

BigX said:
so what is the point in mathematical induction?

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

Joey90 · Jul 23, 2008

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...

kazumi213 · Jul 23, 2008

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.

PizzaPasta · Jul 23, 2008

This thread = YUCK!

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

Elfish · Jul 23, 2008

pwned ^^

CockroachMan · Jul 23, 2008

( 2^n < 3^n )

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

deathfisaro · Jul 24, 2008

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.

Mathematical induction

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