# Having trouble proving stuff

1. Oct 10, 2013

### synkk

Prove for all $x \geq -1$ and n being a natural number, $(1+x)^n \geq 1 + nx$ I've done this using induction

hence deduce from this that for any real number # a \ geq 2 $,$ a^n > n ## for all n belonging to the natural numbers.

I'm stuck at the deduce part

I'm not sure what deduce means, it seems fairly obvious so instead I tried to prove it, by again, induction:

a = 2,
2^n > n which is true a n >= 1
assume true for a = k

then when a = k+1

(k+1)^n > n

I don't see how I can prove it from here, it seems obvious as k >= 2 and n >=1 this will always hold true, and thus a^n > n but I can't seem to prove it. Does anyone have any tips on proving as I'm fairly new at this

thank you

2. Oct 10, 2013

### pasmith

It means "show that this follows from what you have just proved".

In this case, you've shown that if $x \geq -1$ then $(1 + x)^n \geq 1 + nx$.

Now see what happens if you set $a = 1 + x$.

3. Oct 10, 2013

### synkk

how would that work exactly? it states that a >= 2 if we let a = 1 + x then a >= 0 as x >= -1?

Last edited: Oct 10, 2013
4. Oct 10, 2013

### synkk

if I ignore my confusion above I ge tthis:

a^n >= 1 + nx

as x >= -1 a^n >= 1 - n

as n is a natural number n >=1
hence a^n > n