Mathematical Induction (1 Viewer)

[shaon0]

Prove this by mathematical induction. I tried n=1 but it does not satisfy it.

 

[Trebla]

There's a mistake. It should be (3k + 1) rather than (3k - 1)

 

[shaon0]

There's a mistake. It should be (3k + 1) rather than (3k - 1)

Yeah. Thought so.
Thanks for your help.

 

[shaon0]

Prove by mathematical induction:
2^n>=1+n for n>=1
I think i have the solution but i don't know whether it is correct.
Here it is:
Let n=1.
LHS=2.
RHS=2.
Thus, LHS=RHS. (can't be bothered to put all the signs)
Assume; 2^n>=1+n is true for all n>=1.
Let n=n+1.
LHS=2^n+1
>=2^n.2
>=(n+1)*2 [by assumption]
>=2n+2.
Thus, 2n+2>=n+2
Thus, n>=0.
I don't know how to prove this. Need some help thanks.

 

[tommykins]

Assume true for n = k

2^k > 1+k

Prove true for n = k+1

2^(k+1) > k+2
2.2^k > k+2
2(k+1) > k+2
2k + 2 > k+2

2=2 but 2k > k and k > 0

.'. 2k + 2 > k+2

PS. cbf putting equal signs.

 

[lolokay]

yeah I think you just say
2ⁿ >= n+1
2*2ⁿ = 2ⁿ⁺¹ >= 2n + 2 >= (n+1) +1 [since n>0]

I think what you did is pretty much right

 

[shaon0]

yeah I think you just say
2ⁿ >= n+1
2*2ⁿ = 2ⁿ⁺¹ >= 2n + 2 >= (n+1) +1 [since n>0]

I think what you did is pretty much right

Okay. but according to the law of well-ordered pairs. shouldn't nEN(n be an element of Natural numbers)? This also works for all negative numbers.

 

[lolokay]

not sure that I get what you're asking
the induction above doesn't work for negative numbers, since it is specified that n>0 - and the equation only asks for n>= 1

if you mean why the inequality happens to work for all negative numbers, you could solve that by saying when that when n=0,1 then the equation gives equality
and since 2ⁿ is constantly increasing, and at n=0 has a tangent lower than 1 (by differentiating w.r.t n), and at n=1 has a tangent higher than 1, will be >n+1 when n>1, n<0

 

[tommykins]

You're overcmoplicating the question, shaon0.

 

[shaon0]

You're overcmoplicating the question, shaon0.

Okay sorry. I was just wondering.

 

[shaon0]

not sure that I get what you're asking
the induction above doesn't work for negative numbers, since it is specified that n>0 - and the equation only asks for n>= 1

if you mean why the inequality happens to work for all negative numbers, you could solve that by saying when that when n=0,1 then the equation gives equality
and since 2ⁿ is constantly increasing, and at n=0 has a tangent lower than 1 (by differentiating w.r.t n), and at n=1 has a tangent higher than 1, will be >n+1 when n>1, n<0

Thanks, but i won't overcomplicate my question as tommykins said above.
Thanks for your help

 

[tommykins]

Okay sorry. I was just wondering.

Sorry, I took it as something where you questioned the actual answer, didn't know it was your own curiosity.

We know that for any n value, it has to be positive as n > 1

If we assume n = k to be true, we're also assuming k>1

Thus, when you do the induction, it holds as 2k>k where k > 0.

 

[shaon0]

Sorry, I took it as something where you questioned the actual answer, didn't know it was your own curiosity.

We know that for any n value, it has to be positive as n > 1

If we assume n = k to be true, we're also assuming k>1

Thus, when you do the induction, it holds as 2k>k where k > 0.

Okay.