# Entegers

### Entegers

Do not know if these mathematical objects have been explored before but their properties are interesting to me.

They are rational numbers of the form (1 + 1/N) where N is a nonzero positive integer. If they are represented by N and treated as entities unto themselves then we can easily prove that the product of a sequence of these "Entegers" from N = 1 to K is K+1:
1 x2x...K = 1 x 2 x ... K = K+1.

A shorthand for such a sequence can be [N] where it is understood that N starts at 1 goes to N so [N] = N+1

I call these things Entegers because $$\lim_{N\to \infty} N^{N}= e$$ (the base of natural logarithms)

It was interesting to find that any rational number > 1 can be represented by a unique sequence of these things.

Let a, b be positive rational (nonzero) and say b < a so the R = a/b. Since [a-1] = a and [[b-1]b[/b]] = so
[[b]a-1
]/[b-1] = [b, a-1] would be the consecutive sequence.

Have such things been explored before?
Larry

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Joined: Tue Jun 08, 2021 8:35 am
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### Re: Entegers

It is elementary that 1+ 1/n= n/n+ 1/n= (n+1)/n so of course, such a product is
$$\underline{1}\times \underline{2}\times \underline{3}\times\cdot\cdot\cdot\times \underline{N-1}\times\underline{N}$$
$$= (1+ 1/1)(1+ 1/2)(1+ 1/3)\cdot\cdot\cdot (1+ 1/(N-1))(1+ 1/N)$$
$$= 2(3/2)(4/3)\cdot\cdot\cdot(N/(N-1))((N+1)/N$$.

The denominator of each term after the first cancels the numerator of the previous term so that the only terms left are the denominator of the first term, 1, and the numerator of the last term, N+1.
$$\frac{N+1}{1}= N+1$$.
Guest