Question

prove that for every real number e > 0, there is a natural number M such that if m>n>M, then 1/n - 1/m < e

Answer 1

Looks like you're trying to show that the sequence \(a_n=\dfrac{1}{n}\) is a Cauchy sequence?

Answer 2

im honestly not sure, but as a hint in the textbook is says "because m is positive, the statement 1/n - 1/m < e follows from 1/n < e" and when by e it says epislion in the book

Answer 3

Letting \(M=\dfrac{1}{\epsilon}\) should work.

Answer 4

In doing so, you have
\[\begin{align*}
\left|\frac{1}{n}-\frac{1}{m}\right|&=\left|\frac{m-n}{mn}\right|\\
&=\left|\frac{n-m}{mn}\right|&\text{by a property of absolute value}\\
&=\frac{n-m}{mn}&\text{because }m,n>0\\
&<\frac{n}{mn}\\
&=\frac{1}{m}\\
&<\frac{1}{N}&\text{since }m>N\\\\
&=\epsilon
\end{align*}\]

Answer 5

Sorry, that should be an \(M\), not an \(N\)

Answer 6

Okay Thanks, im trying to process this lol.

Answer 7

Is it that some \(n,m\) exist? Or is there something else about \(n\) and \(m\) we should know?

Answer 8

Usually, if you're working this kind of proof out on paper, you'd start by trying to find a proper \(M\) that would work. Think of it in the way you'd write an \(\epsilon-\delta\) proof for a limit.

Answer 9

what i typed in the original response is all that is said in the textbook question and that hint was in the back of the book

Answer 10

So \(n\) and \(m\) don't have to be integers?

Answer 11

so just basically what u did was combined 1/n - 1/m and then since we know that m > n > M, it shows that when combined it is less than the n/nm ( aka 1/m ) and 1/m is less than 1/M

Answer 12

@wio i guess

Answer 13

The fact that \(\epsilon\) is a real number leads me to believe \(m\) and \(n\) are at least positive real numbers. Obviously, they can't be zero, and a proof for negative \(m,n\) would be similar, I think.

And yes @Sobie1337, that's the gist of it.

Answer 14

okay thanks so much. Proofs confuse me so much.... i dont even know why im taking a proof class.

Answer 15

You're welcome!

Answer 16

shouldn't it be (m-n)/(mn) < m/(mn) = 1/n?

because (n-m)/(mn) < 0