proof by induction with inequalities

question is prove by induction on m that m^3 <= 2^m for m=>10

thats supposed to be m cube less than or equal to 2 to the power n for m greater than or equal to 10

P(K)= k^3 less than or equal to 2^k

P(K+1)= (k+1)^3 less than or equal to 2^k+1

2k^3 less than or equal to 2^k+1

(k+1)^3 less than or equal to 2k^3

now im not sure to do next

please help

Re: proof by induction with inequalities

$\displaystyle \begin{align*} (k+1)^3&=k^3+3k^2+3k+1\\ &\le k^3+3k^2+3k^2+k^2&& \text{since }k\ge3 \\&=k^3+7k^2\\&\le k^3+k^3&& \mbox{since }k\ge7\\&\le2^k+2^k&& \text{by induction hypothesis}\\&=2^{k+1}\end{align*}$

A couple of remarks. For each claim it should be clear whether you are assuming it, trying to prove it or something else. Your proof is just a series of statements, and the role of those statements is not clear. Write comments when necessary, such as "We assume ...," "We need to prove ...," "It follows from ... that ..."

Mathematics is case-sensitive, so K and k may denote different variables.

Edit: Don't forget the base case.

Re: proof by induction with inequalities

Quote:

Originally Posted by

**emakarov** $\displaystyle \begin{align*} (k+1)^3&=k^3+3k^2+3k+1\\ &\le k^3+3k^2+3k^2+k^2&& \text{since }k\ge3 \\&=k^3+7k^2\\&\le k^3+k^3&& \mbox{since }k\ge7\\&\le2^k+2^k&& \text{by induction hypothesis}\\&=2^{k+1}\end{align*}$

thanks but i dont understand where the second line has come from, how did you get k^3+3k^2+3k^2+k^2

Re: proof by induction with inequalities

Quote:

Originally Posted by

**narjsingh** thanks but i dont understand where the second line has come from, how did you get k^3+3k^2+3k^2+k^2

$\displaystyle 1\le k^2$

Re: proof by induction with inequalities

Quote:

Originally Posted by

**Plato** $\displaystyle 1\le k^2$

im really really sorry i still dont get that im a bsc econ student :(

Re: proof by induction with inequalities

Quote:

Originally Posted by

**narjsingh** im really really sorry i still dont get that im a bsc econ student :(

Because $\displaystyle k\le k^2~\&~1\le k^2$ we get $\displaystyle 3k+1\le 3k^2+k^2$

$\displaystyle k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2$

Re: proof by induction with inequalities

Quote:

Originally Posted by

**Plato** Because $\displaystyle k\le k^2~\&~1\le k^2$ we get $\displaystyle 3k+1\le 3k^2+k^2$

$\displaystyle k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2$

thank you so much! i am really getting it! one last hing how did you get k is greater than or equal to 7? :)

Re: proof by induction with inequalities

Quote:

Originally Posted by

**narjsingh** thank you so much! i am really getting it! one last hing how did you get k is greater than or equal to 7? :)

$\displaystyle 3+3+1=7$ so $\displaystyle 3k^2+3k^2+k^2=7k^2$