Question

Prove
For every real number A>0, there is a natural number B such that for all real numbers C>B,
1/(4C) 13 years ago

Answer 1

Let's see here.

Case 1: \(A\geq1\). If you take \(B=A\) then you have that \(4C\geq4A\). So you have that \[\frac{1}{4C}\leq\frac{1}{4A}\]Since \(A\geq1\), \(\frac{1}{A}\leq A\) with equality when \(A=1\), so \[\frac{1}{4C}<A\]

Now we need a case 2 with \(A<1\).

Answer 2

there is something wrong with your question ! you can't have for all real numbers 1/(4C)<A

Answer 3

am I right KingGeorge ?!

Answer 4

Case 2: \(A<1\). Now take \(B=\lceil \frac{1}{A}\rceil\). In this case, \(C>\frac{1}{A}\) so \[\frac{1}{C}<A\]Which implies \[\frac{1}{4C}<A\]

I'm pretty sure this statement is in fact, true.

Answer 5

Yes it is true

Answer 6

Note: In case 2, you should take \(B=\lceil \frac{1}{A}\rceil +1\) to ensure that \(C>\frac{1}{A}\).

Answer 7

okk gotcha still need to reread the proof to process it

Answer 8

I'm sure it can't be true ! you can't choose the C !

Answer 9

sorry ! I didn't understand this correctly :D ! for all real nnumers > B ...we have.....
sorry...for wasting your time :D

Answer 10

Ya im like that too i read it too fast and totally misread it

Answer 11

Dont say u r like that KG cuz u r not

Answer 12

It took me a few minutes to decide if this was true or false. It's certainly not intuitive.

Answer 13

Thanks for helpin me out :D

Answer 14

No problem

Answer 15

umm King can i ask u a question?

Answer 16

Y did u have to split it up into 2 cases where A>1 and where A<1

Answer 17

You might not have to. It was part of my process of determining if it was actually true or not.

However, the argument for case 1 doesn't work for case 2, the argument for case 2 doesn't work for case 1. So if we didn't split it up into two cases, we would need an entirely different argument.

Answer 18

well lets say I wld say b=(1/(2a)). i mean it cld be niething

Answer 19

If \(B=\lceil\frac{1}{4A}\rceil+1\), then \(C>\frac{1}{4A}\) so we know that \[4C>\frac{1}{A}\implies\frac{1}{4C}<A\]That seems to work.

Answer 20

okkkk yupppp

Answer 21

If we choose \(A=10\) then \(B=2\), so \(C>2\). So \[\frac{1}{4C}<\frac{1}{8}<A\]So that seems to work.

Answer 22

okkkk gotcha thanks

Answer 23

you're welcome.