Need help /w a math problem

[Chu]

Can anyone tell me what the equality symbol in this problem means?

link

 

[HomeBrewerDude]

Originally posted by: Chu
Can anyone tell me what the equality symbol in this problem means?

link

this should go in the homework forum.

 

[Chu]

Originally posted by: HomeBrewerDude

Originally posted by: Chu
Can anyone tell me what the equality symbol in this problem means?

link

this should go in the homework forum.

Except it's not homework . . .

 

[blahblah99]

Originally posted by: Chu
Can anyone tell me what the equality symbol in this problem means?

link

Probably means less than or equal to whenever you see that symbol, or greater than or equal to. I think it's just a lazy way of writing concatenating two problems into one.

 

[dullard]

To me it appears like 3 problems. Prove it when using a '>' sign, then prove it when using an '=' sign, and then prove it when using the '<' sign.

I could be wrong though.

 

[raptor13]

It's a cases problem. You have to prove the first statement when k takes any of three values... <, =, or >.

Sucks to be you!

 

[TuxDave]

Yeah, prove it for both 'less than' case, prove it for both 'equal' case and prove it for both 'greater' case

 

[MrSmithers]

I agree, 3 seperate problems. OTOH, what the hell kind of English is "according as?" I would call total shenanigans on the whole problem based merely on that.

Smithers

 

[So]

 

[CSMOOTH]

Here is my thoughts on it... Take it as you will:
The rest is trivial...

So if you expand the nCr on both sides you get:
n! / [ (k-1)! * (n-(k-1))! ] ? n!/[k! * (n-k)!]

the numberators are the same so if the denominator is the greater on the left then the value to the right is larger (sign inversion). eg. 1/6 < 1/5

so you have
(k-1)! * (n-(k-1))! ? k! * (n-k)!

substitute
K! = (K-1)! * K on the right side and you have
(k-1)! * (n-(k-1))! ? (K-1)! * K * (n-k)!

Divide through by (k-1)!
(n-(k-1))! ? k * (n-k)!

substitute n-(k-1) = n-k+1
(n-k+1)! ? k * (n-k)!

substitute (n-k+1)! = (n-k)! * (n-k+1)
(n-k)! * (n-k+1) ? k * (n-k)!

divide by (n-k)!
n-k+1 ? k

add k to both sides
n+1 ? 2k

rewritten:
k ? (n+1) / 2

-Smooth