How would I prove the following: If a and b are real numbers such that a 1 + (k+1)a

Question

How would I prove the following:

If a and b are real numbers such that a<b, then -b<-a?

and

if a is an element of R, and a does not equal 0, and a is greater than -1, then for all k that are elements of N, (1+a)^(k+1) > 1 + (k+1)a

usukidoll · Answer

isn't one of those questions Bernoullis?!

anonymous · Answer

im not sure what that is.. my professor gave us some exercises to work and im having trouble with those 2

usukidoll · Answer

one of them looks familiar... like http://en.wikipedia.org/wiki/Bernoulli%27s_inequality

anonymous · Answer

im not sure about that..  How would I prove it though?

usukidoll · Answer

idk...I'm in a proof writing class myself and my class is going crazy

anonymous · Answer

lol same here.. .-.

anonymous · Answer

if a < b, then by definition of <, b + addin(a)  ∈ R+

let n = b + addin(a)

n = n + 0 by additive identity 
n = n + b + addin(b),  by additive inverse
n = n + addin(addin(b)) + addin(b), by the lemma addin(addin(n)) = n
n = b + addin(a) + addin(addin(b)) + addin(b), by substitution 
n = b + addin(b) + addin(a) + addin(addin(b)), by associative 
n = 0 + add(a) + addin(addin(b)), by additive inverse
n = add(a) + addin(addin(b)), by identity property

and since n∈ R+, and by definitio of <, 
add(b) < add(a)

which means -b < -a 

man I suck -.-

usukidoll · Answer

Here's a great question. W T Fuu fuu invented pure math?

anonymous · Answer

humans XD

anonymous · Answer

I believe you can use binomial theorem for the second proof

usukidoll · Answer

oh no! X( torture

usukidoll · Answer

use induction XD

anonymous · Answer

yep, believe I got it

anonymous · Answer

base case k = 1
(1+a)^(1+1) >? 1 + (1+1)a
(1+a)^2 >? 1 + 2a
1 + 2a + a^2 > 1+2a (check)

suppose (1+a)^(k+1) > 1 + (k+1)a for for some integer k,

consider (1+a)^(k+1+1) = (1+a) (1+a)^(k+1)

by hypothesis:
(1 + a)^(k+1) > 1 + (k+1)a

since a > -1, then a+1 > 0, so we can multiply both sides by (1+a) without changing the sign
(1+a) (1+a)^(k+1) > [1 + (k+1)a ] (1+a)

expand the right side,
> 1+ 2a + a^2+ ak + a^2 k

> 1 + 2a + ak

> 1 + (a + a + ak) = 1 + (1 + 1 + k) a

check XD