mathematical induction (1 Viewer)

[samuelclarke]

Prove by mathematical induction:

2^n greater than or equal to n^2 for all integers n greater than or equal to 4

thanks!

 

[Carrotsticks]

I'll skip the n=4 step and the conclusion, since those are trivial.

Hypothesis:



Inductive step k --> k+1:

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=The~statement~2^{n}\geqslant n^2~for~n\geqslant 4.\\ Let~n=4~to~show,\\ LHS=2^4=16\\ RHS=4^2=16\\ LHS\geqslant RHS~\therefore it~is~true~for~n=4\\ \\ Assume~true~for~n=k~where~k~is~an~integer\geq 4\\ i.e.~2^k\geq k^2\\ \\ Use~this~to~prove~true~for~n=k@plus;1\\ TBP:~2^{k@plus;1}\geq (k@plus;1)^2\\ LHS=2^{k@plus;1}\\ =2 . 2^k\geq 2k^2(from~above)\\ =(k@plus;1)^2-2k@plus;k^2-1\\ =(k@plus;1)^2@plus;(k-1)^2-2\geq (k@plus;1)^2,(as~ k-1\geq 3,~and~(3)^2\geq 2)\\ \\ If~it~is~true~for~n=k,~then~it~is~true~for~n=k@plus;1.~As~it~is~true~for~n=4,~it~will~be~true~for~n=5,~and~so~on.~Therefore,~by~the~laws~of~mathematical~induction,~it~is~true~for~all~n\geq 4." target="_blank"><img src="http://latex.codecogs.com/gif.latex?The~statement~2^{n}\geqslant n^2~for~n\geqslant 4.\\ Let~n=4~to~show,\\ LHS=2^4=16\\ RHS=4^2=16\\ LHS\geqslant RHS~\therefore it~is~true~for~n=4\\ \\ Assume~true~for~n=k~where~k~is~an~integer\geq 4\\ i.e.~2^k\geq k^2\\ \\ Use~this~to~prove~true~for~n=k+1\\ TBP:~2^{k+1}\geq (k+1)^2\\ LHS=2^{k+1}\\ =2 . 2^k\geq 2k^2(from~above)\\ =(k+1)^2-2k+k^2-1\\ =(k+1)^2+(k-1)^2-2\geq (k+1)^2,(as~ k-1\geq 3,~and~(3)^2\geq 2)\\ \\ If~it~is~true~for~n=k,~then~it~is~true~for~n=k+1.~As~it~is~true~for~n=4,~it~will~be~true~for~n=5,~and~so~on.~Therefore,~by~the~laws~of~mathematical~induction,~it~is~true~for~all~n\geq 4." title="The~statement~2^{n}\geqslant n^2~for~n\geqslant 4.\\ Let~n=4~to~show,\\ LHS=2^4=16\\ RHS=4^2=16\\ LHS\geqslant RHS~\therefore it~is~true~for~n=4\\ \\ Assume~true~for~n=k~where~k~is~an~integer\geq 4\\ i.e.~2^k\geq k^2\\ \\ Use~this~to~prove~true~for~n=k+1\\ TBP:~2^{k+1}\geq (k+1)^2\\ LHS=2^{k+1}\\ =2 . 2^k\geq 2k^2(from~above)\\ =(k+1)^2-2k+k^2-1\\ =(k+1)^2+(k-1)^2-2\geq (k+1)^2,(as~ k-1\geq 3,~and~(3)^2\geq 2)\\ \\ If~it~is~true~for~n=k,~then~it~is~true~for~n=k+1.~As~it~is~true~for~n=4,~it~will~be~true~for~n=5,~and~so~on.~Therefore,~by~the~laws~of~mathematical~induction,~it~is~true~for~all~n\geq 4." /></a>

Can others tell me if there could be any improvement in my proof?

 

[math man]

I'll skip the n=4 step and the conclusion, since those are trivial.

your step is probably too complicated for them, probably best to do it like:

 

[math man]

although i do prefer your method carrot

 

[asianese]

Your conclusion can be shortened - fact from BOS teacher's meeting:

Hence proved by mathematical induction true for n >= 4. Is sufficient nowadays. The long winded - if it's true for k, then k+1 then blah blah - isn't needed.

 

[Carrotsticks]

Your conclusion can be shortened - fact from BOS teacher's meeting:

Hence proved by mathematical induction true for n >= 4. Is sufficient nowadays. The long winded - if it's true for k, then k+1 then blah blah - isn't needed.

Here is my conclusion, which is even faster than that.

 

[math man]

now thats my kind of conclusion

 

[iSplicer]

Here is my conclusion, which is even faster than that.

Suave. So suave!

Nah but seriously, you're a legend carrotsticks. Very impressed by the elegance of your solutions and your willingness to help people out. Well done!

 

[samuelclarke]

oh cool thanks for the help!