Maths problem...

[Nineshadow]

Demonstrate that for every real k >= 2 :

 

                     and                  (√k - √(k-1))/k  < 1/√(k-1) - 1/√k

 

 

 

Damn. :l

[nanowatt]

Put some picture i can't understand anything

[Nineshadow]

Put some picture i can't understand anything

Check original post.

[soup]

Ignore shouldn't post before coffee.

 

Mods pls delete.

[Archangel1994]

The first one is relatively simple in method. On the right side of the > You need to give both terms a common denominator so they can be subtracted from each other to make 1 fraction to the left of the > and and 1 fraction to the right of it. Then you can do Cross-fraction multiplication (top left * bottom right) = (Bottom left * top right) and you should be able to get the answer from there.

 

The second one appears to be the same thing but because that is what you TYPED out it's a bit hard to see. Both 1 and 2 seem to consist of 3 fractions though. You need to simplify both to 2 fractions with 1 on each side of the < or > and then do Cross-fraction multiplication so you end up with non-fraction terms on each side of the < or >. From there on it's a relatively simple task to isolate k

[Nineshadow]

The first one is relatively simple in method. On the right side of the > You need to give both terms a common denominator so they can be subtracted from each other to make 1 fraction to the left of the > and and 1 fraction to the right of it. Then you can do Cross-fraction multiplication (top left * bottom right) = (Bottom left * top right) and you should be able to get the answer from there.

 

The second one appears to be the same thing but because that is what you TYPED out it's a bit hard to see. Both 1 and 2 seem to consist of 3 fractions though. You need to simplify both to 2 fractions with 1 on each side of the < or > and then do Cross-fraction multiplication so you end up with non-fraction terms on each side of the < or >. From there on it's a relatively simple task to isolate k

It'll end up like :

1 > k , which is false. :l (first one)

[Archangel1994]

It'll end up like :

1 < k , which is false. :l (first one)

I have received these kinds of questions before and worked them out like this and it always worked for me. Maybe you went wrong somewhere?

[soup]

 

1 < k , which is false. :l (first one)

No it's not, if k is greater than 2 (further from 0 in the set of real numbers) it is obviouslly greater than 1.

So not false just not the answer you were looking for.

[Nineshadow]

 

I have received these kinds of questions before and worked them out like this and it always worked for me. Maybe you went wrong somewhere?

 

This isn't a calculus problem.It's an inequality,a bitch of a problem :l.

[kingkickolas]

Demonstrate that for every real k >= 2 :

 

                     and                  (√k - √(k-1))/k  < 1/√(k-1) - 1/√k

 

 

 

Damn. :l

Well the first inequality is self-evidently not true. Observe the case for k = 2.

In fact, it looks like the inequality should be the other way around. In which case, algebra will give you "k > 1", which completes the proof.

[kingkickolas]

Demonstrate that for every real k >= 2 :

 

                     and                  (√k - √(k-1))/k  < 1/√(k-1) - 1/√k

 

 

 

Damn. :l

The second proof might look something like this: