Basically, a function is a formula. To be a bit more specific, the simplest type of a function is a rule that takes a single number and gives a unique, unambiguous number as an output. A function is just a machine for generating a number from a number. (More sophisticated functions are rules involving vectors, but it is still a formula or rule.)
f(x) = 2x + 1 means to take the number x, whatever it is, as input and spit out twice x plus 1 as the output. If you click on the word function shown in blue in the quotation of your most recent post, you will get a more complete explanation of a function.
Do you know the symbol \(\displaystyle \sum\), which means to sum things up? A large capital sigma, the Greek equivalent to "S", stands for sum.
The symbol \(\displaystyle \prod\) means to multiply things. A large capital pi, the Greek equivalent to "P", stands for product.
\(\displaystyle So\ \prod{i}\ for\ i = 1, ...\ 5 = 1*2*3*4*5 = 120\)
It is probably not necessary to substitute m = (n / 2). The set of the first 100 positive even integers is identical to the set of twice the first 50 positive integers. It seems slightly easier to me to set up the function in terms of m successive integers than n integers not in succession. I am very lazy and do things the easiest way I can find. In any case, doing it in terms of m relates this problem very directly to Euclid's proof that there is no largest prime and that the number of primes is not finite.
None of this is essential to solving your problem. The first important thing is whether you see that
\(\displaystyle If\ m = \frac{n}{2}, h(100) = g(50) = 1 + \prod{(2i)}\ for\ i = 1, ... 50 =\) \(\displaystyle 1 + [(2*1)*(2*2) * ...\ (2 * 50)]\)
expresses in symbols what the problem describes in words. You good there?
Now consider an arbitrary prime, \(\displaystyle 1 < p \leq 50\) (greater than 1 because 2 is the smallest prime).
\(\displaystyle Let\ q = \dfrac{g(50)}{p}\)
\(\displaystyle Let\ r = \dfrac{\prod{(2i)}}{p}\ for\ i = 1, ...\ 50\)
\(\displaystyle So\ q = r + \dfrac{1}{p} \)
Obviously (1 / p) < 1 is not an integer. So if r is an integer, q = r + (1 / p) is not an integer. If I add a fraction to an integer, I do not get an integer.
Question: is r an integer, and if so why?
Can you complete the problem now? We are at the last step.
Edit The big pi and sigma signs are also shown as \(\displaystyle \prod{(2i)}\ for\ i = 1, ...\ 50 =\prod_{i=1}^{50}(2i) =\displaystyle \prod_{i=1}^{50}{(2i)}\)
These big pi and big sigma symbols are exact and concise so they let you see accurately on paper what you are trying to think about.
