Problem 44 - Page 2

Henderson · #11 $Old$ 01-27-2008, 09:42 AM

Let $A_{n} = 1+\frac{1}{2}+...+\frac{1}{n}$ .

Assume that, for some $n$ , $A_{n}$ is an integer.

Let $P$ be the product of all the denominators except for the largest prime less than $n$ . $P$ is clearly an integer.

Then $P \cdot A_{n} = P \cdot (1+\frac{1}{2}+...+\frac{1}{n})$ .

On the LHS, $P \cdot A_{n}$ is an integer, based on how we defined P and our assumption.

On the RHS, after distributing, each term becomes an integer except for the term with the largest prime as the denominator. This clearly is not an integer, so the RHS is the sum of many integers plus one non-integer, and the RHS as a whole is not an integer.

Left with the integer LHS equaling a non-integer RHS, our assumption must be false, and $A_{n}$ must not be an integer.

$Q.E.D.$

Skinner · #12 $Old$ 01-31-2008, 10:14 PM

Let me have a try:

problem:

1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Solution:

If you can prove that $(\frac{1}{2} + ... + \frac{1}{n})$ is not an integer, then you can prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Why? if you add 1 to anything that's not an integer, you get a non-integer value.

look at this function:

$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$

This rational function has asymptopes at the integers, which means that at every integer, the function approaches it, but never reaches it. This can be proven if we took the limit of the function as $x \rightarrow 1+, 2+, ....$ and so on. This means that the value of $f(n)$ never becomes an integer.

I conclude that $f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2} = (\frac{1}{2} + ... + \frac{1}{n})$

$1 + f(n)$ is not an integer, therefore $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer either.

Please tell me what you think.

janvdl · #13 $Old$ 02-01-2008, 05:46 AM

Quote:

Originally Posted by Skinner $View Post$

Let me have a try:

problem:

1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Solution:

If you can prove that $(\frac{1}{2} + ... + \frac{1}{n})$ is not an integer, then you can prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Why? if you add 1 to anything that's not an integer, you get a non-integer value.

look at this function:

$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$

This rational function has asymptopes at the integers, which means that at every integer, the function approaches it, but never reaches it. This can be proven if we took the limit of the function as $x \rightarrow 1+, 2+, ....$ and so on. This means that the value of $f(n)$ never becomes an integer.

I conclude that $f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2} = (\frac{1}{2} + ... + \frac{1}{n})$

$1 + f(n)$ is not an integer, therefore $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer either.

Please tell me what you think.

You'll have to see what TPH thinks, but if your proof is right, i say Well Done! It's very easy to understand as well

ThePerfectHacker · #14 $Old$ 02-01-2008, 10:21 AM

Quote:

Originally Posted by Skinner $View Post$

$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$

How is this even a function? What is the value of $f(1/2)$ ? In order to talk about asymptotes you need to have a function defined at every single point (including $1/2$ ) except at the asymptote points.

Skinner · #15 $Old$ 02-01-2008, 02:44 PM

oh. I messed up. Fail.

Jameson · #16 $Old$ 02-01-2008, 11:40 PM

Quote:

Originally Posted by Skinner $View Post$

oh. I messed up. Fail.

Good try though! Don't be discouraged. I have failed much worser doing far easier proofs.

Aryth · #17 $Old$ 02-06-2008, 02:43 PM

The Sum:

$\sum_{i = 1}^{k}\frac{1}{k}$

Does not yield an integer for all $k \geq 2$ .

For k = 2:

$1 + \frac{1}{2} = \frac{3}{2}$

That is not an integer.

Since the series is divergent, then we must reduce it:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right]$

To have an integer it has to be that for some integer M:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right] = M$

$(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1) = (2*3*...*k)*M$

This can be rewritten as:

$k*((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) + (2*3*...*k-1) = (2*3*...*k)*M$

We can say that:

$((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) = C$

So that:

$k*C + (2*3*...*(k-1)) = (2*3*...*k)*M$

We can also say that:

$(2*3*...*(k-1)) = N$

So that our final equation is:

$kC + N = (2*3*...*k)M$

The series is known as the Harmonic series, and is often dealt with for primes.

If k is prime, this is the sum of a multiple and a non-multiple of k.

Clearly there is no M that satisfies the equation.

The Harmonic Series does not give an integer for any $k\geq2$

$Q.E.D.$

bjhopper · #18 $Old$ 02-10-2008, 09:04 PM

two trains 100 miles apart are approaching at 30 miles per hour.they will collide after 3.33 hrs. the speed of the faster must be less than 30mph, the speed of the slower greater than zero neglect not moving.the bird wings it for 3.33 hrs so travels 200 mi . 60 x3.33hrs.
example train A 20 mph, train B 10 mph. they aapproach each other at 10 plus 20 mph .train Atravels 67 mi B 33mi