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1. how would you prove that for all integers > 0, and given that x>=-1:

(1+x)^n >= 1 + nx

i tried doing this by induction....it's true for n=1

so suppose (1+x)^k >= 1 + kx
then:

(1+x)^(k+1) = (1+x)((1+x)^k)

so: (1+x)^(k+1) >= (1+x)(1+kx)
>= 1 + (k+1)x + kx^2

but how do you lose the kx^2?
2. Simple, x^2 >= 0, so
1 + (k+1)x + kx^2 >= 1 + (k+1)x
3. doh yeah of course!
4. so hold on....is the condition x>=-1 actually required?
5. this is correct but should really use whole binomial expansion and then use x^n is always greater than or equal to zero if x>0 (this is easily shown inductively:

x^k > 0
x^(k+1) = x (x^k)

Since x is positive or equal to zero . this must also be greater than or equal to zero.

p.s. this is only strictly true if n and x are integers
6. bionomial exapnsion not allowed to be used in answer i'm afraid!
7. no see i havent involved the x>=-1 condition, so the proof isnt complete yet! how do I involve that part!?
8. (1+x)^n >= 1 + nx

(1+x)^0 = 1
1 + 0x = 1

so it ss true for n=0

assume tru for all n>0
(1+x)^n >= 1 + nx

(1+x)^(n+1) = (1+x)(1+x)^n

1 + (n+1)x = 1 + nx + x

so is this true?:
(1+x)(1+x)^n >= 1 + nx + x

from our assumption we know that (1+x)^n >= 1 + nx.
however the modification is we multiply the LHS by x+1 and add x to the RHS. as x>0, and for n>0 (1+x)^n > 1, the LHS will always be larger than the LHS.

so it is true for all n.

I think that should work?
I just did the induction on n rather than x.
9. Let P(n) be the statement that (1+x)^n >= 1 + nx

Base Step

Check that P(n) is true for n = 1:

(1 + x)^1 >= 1 + x (which is true)

Inductive Step

Assume P(n) is true for all n > 0. Then P(n+1) is:

(1 + x)^(n+1) = [(1 + x)^n].(1 + x)

But you've assumed P(n) is true, so

(1 + x)^(n+1) >= (1 + nx).(1 + x)
>= 1 + x(n+1) + n.x^2

As x >= -1 and n > 0 (this is where the condition of x plays the part in your proof), the expression n.x^2 will always be equal or greater than 0.

so original statement >= 1 + x(n+1)

Showing that P(n) true => P(n+1) true and we are done QED

Euclid.

Edit: (Sorry if I have repeated any thing from above, as I went straight into this reply and didn't read the previous posts)
10. (Original post by Euclid)

As x >= -1 and n > 0 (this is where the condition of x plays the part in your proof), the expression n.x^2 will always be equal or greater than 0.
i dont think the fact x>-1 implies n.x^2>0 for any n>0 this will always be true.
the x>=-1 ensures 1+x>=0
so if we assume (1+x)^k>1+kx then if we multiply by (1+x) for the induction step the inequality changes since (1+x)<0.
11. (Original post by evariste)
i dont think the fact x>-1 implies n.x^2>0 for any n>0 this will always be true.
the x>=-1 ensures 1+x>=0
so if we assume (1+x)^k>1+kx then if we multiply by (1+x) for the induction step the inequality changes since (1+x)<0.
Hmm yes, I'm starting to become really sloppy in my work. I should really think before saying anything .

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Updated: January 16, 2005
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