Hello i need help with this proof VERY BADLY!


Prove or disprove, as appropriate, each of these statements. You may use all the proof techniques we’ve used so far: direct proof, existential instantiation, generalizing from generic particular, counterexamples, contradiction, contraposition, etc., as well as basic algebra.

1. Prove or disprove that:

for integers n that are equal to or greater than 2, square root of n is less than [(1/(square root 1)) + (2/(square root 2)) + (n/(square root n))]

i recommend proof by induction

Prove for the base case (n=2)
then assume n=k for all k in the Integers k >1
then prove its true for (k+1)

This post was edited by TritonV8 on Apr 19 2014 09:13pm

Are you positive one the question ?

Just asking because :

(n/(square root n)) appears to be square root (n)

hence, the result is ridiculously obvious...

[(1/(square root 1)) + (2/(square root 2)) + (n/(square root n))] = 1 + root(2) + root(n) just by simplifying the radicals
so....
root(n) =?= 1+root(2)+root(n)
0 =?= 1 + root(2) by subtraction of root(n) on both sides
0 < 1 + root(2)

this works for all n, this doesnt even have to be for integers greater than or equal to 1. make sure u copied the question down EXACTLY right, this seems like a retarded question to give out if that's really all they give u
im also concerned about the same part of what feanur said, u have a root(n) on both sides of the equal sign which just cancels out when u subtract it on both sides, leaving you with just numbers and no variables

5/(sq root 5) is not the same as just square root 5

but in the case that n=1: 1/(sq root 1) is just 1

wrong, 'feanur' is right since n = sqrt(n) * sqrt(n)

so the op is trivial except if only integer arithmetic is to be used, ie n/sqrt(n) in [...] denotes the integer less equal to the real number n/sqrt(n)

This post was edited by brmv on Apr 20 2014 05:27pm

This is what I need to do, unless someone can prove it easier via algebra

but yes the problem is to prove/disprove that for integers n that are equal to or greater than 2, square root of n is less than [(1/(square root 1)) + (2/(square root 2)) + ... (n/(square root n))]

like fenur said, it's trivial. i assume you can do your base case just fine.

what you're left with is:

rt(n) <= sum(rt(i) for i = 1 to n)
rt(n+1) <= sum(rt(i) for i = 1 to n) + rt(n+1)
now cancel out rt(n+1) from both sides
0 <= sum(rt(i) for i = 1 to n)

and we know the sum on the right is positive, so we know it's true.