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:

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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 [imath]n = 2^k[/imath].

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 [imath]n = 2^k[/imath].

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 [imath]n + 1 < 2^{n+1}[/imath] 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 \(\displaystyle h(-\infty) = 2^{-\infty} = 0\)

 

[logistic_guy]

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

Let us check the base case again.
If \(\displaystyle 1 < 2^1\)
I will assume that this is true for any arbitrary \(\displaystyle k\). That is:

\(\displaystyle k < 2^{k}\)

Then, I need to show that:

\(\displaystyle k + 1 < 2^{k + 1}\)

I need to do any tricks on \(\displaystyle k < 2^{k}\) to show that the above inequality is true. Let us try by adding \(\displaystyle 1\) to both sides.

\(\displaystyle k + 1 < 2^{k} + 1\)

I am confident that \(\displaystyle 2 + 1^n < 2 + 2^n\), then

\(\displaystyle k + 1 < 2^{k} + 1 < 2^{k} + 2^{k} = 2^{k+1}\)

Then,

\(\displaystyle k + 1 < 2^{k+1}\)