induction

[logistic_guy]

Use mathematical induction to prove that $\displaystyle n < 2^n$ whenever $\displaystyle n$ is a positive integer.

 

[khansaheb]

Use mathematical induction to prove that $\displaystyle n < 2^n$ whenever $\displaystyle n$ is a positive integer.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

Let us test some values.

$\displaystyle 1 < 2^1 = 2$
$\displaystyle 2 < 2^2 = 4$
$\displaystyle 3 < 2^3 = 8$

It seems that the inequality holds for any positive integer $\displaystyle n$.

@khansaheb

What's next?

 

[khansaheb]

mathematical induction

Please define the term displayed above - as it is presented in your textbook

 

[logistic_guy]

Please define the term displayed above - as it is presented in your textbook

Thank you Sir khan.

 

[khansaheb]

@khansaheb

What's next?

Try reductio ad absurdum.

 

[logistic_guy]

Try reductio ad absurdum.

Could you please show me what do you mean by this?

 

[khansaheb]

Could you please show me what do you mean by this?

What do you find if you google the term "reductio ad absurdum"?

 

[logistic_guy]

What do you find if you google the term "reductio ad absurdum"?

I don't use google or any other technology in the beginning of the attack. I first attempt to solve the problem with my current knowledge and ideas. When I get deeply stuck I look at my references and books. After that when I am totally lost, Mr. google might be useful.

For now it's either you help me and explain what you mean or it was a lie when you told me where I was stuck!

 

[khansaheb]

I don't use google or any other technology in the beginning of the attack.

Well then use your text-book/class-notes and find the definition of that Latin term.

 

[logistic_guy]

Well then use your text-book/class-notes and find the definition of that Latin term.

Thank you again Sir khan.

 

[Agent Smith]

Let $n = 2^k$.

Running an old module. Looks like it might work

 

[logistic_guy]

Agent Smith is here

Where have you been man? We missed your $\displaystyle \infty$ topics

Do you agree $\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$?

And tell us frankly what was the real reason behind your absence for months! Have you got married to Latina or something?

Let $n = 2^k$.

Running an old module. Looks like it might work

You meant $\displaystyle n = 2^n$, right?
I have been told that if I can show $n + 1 < 2^{n+1}$ then the proof is complete by induction.

But $\displaystyle n = 2^n$ is never true because you can think of it as two functions. Say we have $\displaystyle g(x) = x$ and $\displaystyle h(x) = 2^x$. We have been told that the exponential function grows faster and never becomes zero.

Are they lying to us? Because $\displaystyle h(-\infty) = 2^{-\infty} = 0$