Prove by induction that for any x, if x >1, then 3n >1 + 2n.?

Answer 1

What is the relationship between n and x in your problem?

If you meant n > 1 instead of x > 1
then your inequality can be written n > 1
And we wish to prove this by induction.

OK. Start with n = 2.
We verify that 2 > 1

Then assume that it is true for some n = r
ie r > 1
Is it also true for r+1?
Well since r > 1 it is positive
therefore r + 1 > r > 1
thus r+1 > 1

Proved by induction.
That question was kinda stupid, don't you think?

Answer 2

Induction, I believe, means you test it out for n (or x) =2 (the lowest value), then show that the same thing is true for n+1 (or x+1)...

So, for n=2, 3*2 = 6, which IS greater than 1+ 2*2=5..

Now if you plug in (n+1) in the place of n...

3(n+1) > 1 + 2(n+1)
3n + 3 > 1 + 2n + 2

Now group the original equation up to get:

(3n) + 3 > (1+2n) + 2

Since 3 > 2 (the "extra" stuff), this completes the proof by induction that for n>1, 3n>1+2n.

I think it goes something like that...

Answer 3

Assume that P(k) is true where P(k) is proposition that :-
3k > 2k + 1 for k > 1

Consider P(2):-
LHS = 6
RHS = 5
6 > 5 thus P(2) is true.

Consider P(k + 1):-
3(k + 1) > 2(k + 1) + 1
3 (k + 1) > 2k + 3------this is P(k + 1)

Now , starting with P(k) which is true:-
3k > 2k + 1
3k + 3 > 2k + 1 + 3
3 (k + 1) > 2k + 4 > 2k + 3
3(k + 1) > 2k + 3
Thus P(k+1) is true

Since P(2) is true and P(k + 1) is true then
P(k) is true.
Thus 3k > 1 + 2k is true and therefore
3n > 1 + 2n is true for n > 1