I was reading through Mathematical Foundations of computing (preliminary course note by keith Schwarz pg 15 ) and noticed there was a definition for sets of positive natural numbers $\mathbb{N}^+ = \{ 1,2,3…\}$, noting that $0 \notin \mathbb{N} ^+ $ and I also read in Mathematics for computer science ( by Eric Lehma pg 14) that $\mathbb{Z}^+$ represents positive integers having the same set $ \{1,2,3,\dotsc\}$. So my question is, are they the same set? Is there any difference between the set of positive natural numbers and positive integers?
Difference between $\mathbb{Z}^+$ and $\mathbb{N}^+$
integersnotation
Related Solutions
No. The statement is still true. The cardinality of the natural number set is the same as the cardinality of the rational number set. In fact, this cardinality is the first transfinite number denoted by $\aleph_0$ i.e. $|\mathbb{N}| = |\mathbb{Q}| = \aleph_0$. By first I mean the "smallest" infinity.
The cardinality of the set of real numbers is typically denoted by $\mathfrak{c}$. We have $\mathfrak{c} > \aleph_0$, since we can set up a bijection from $\mathbb{R}$ to the power set of the natural numbers and by Cantor's theorem, for any set $X$, we have $|X| < |2^{X}|$. So we have $|\mathbb{R}| = |2^{\mathbb{N}}| > |\mathbb{N}|$. So what this essentially says is that "there are more real numbers (which include rational and irrational numbers) than there are integers" in some sense.
The continuum hypothesis states that "there is no set whose cardinality is strictly between that of the natural numbers and that of the real numbers" which essentially means real numbers form the second "smallest" infinity.
First of all, kudos for introducing these ideas to your 10 year-old! That's an excellent way to get them interested in mathematics at an early age.
As to your question: The short answer is no. Any algebraic operation that you can do of the sort you're describing will yield a complex number. This is due to the fact that they are algebraically closed. What this means is the following.
One way that you can show that you can find a larger domain than the real numbers is by looking at the polynomial $$ f(x) = x^2 + 1 $$ You can easily see that there are no solutions to the equation $f(x) = 0$ in the real numbers (just as the equation $x + 1 = 0$ has no solutions in the positive integers). So we must escape to a larger domain, the complex numbers, in order to find solutions to this equation.
A domain (to use your term) being algebraically closed means that every polynomial with coefficients in that domain has solutions in that domain. The complex numbers are algebraically closed, so no matter what polynomial-type expression that you write down, it will have as solution a complex number.
Now, this isn't to say that there aren't larger domains than $\mathbb{C}$! One example is the Quaternions. Where the complex numbers can be visualized as a plane (i.e. $a + bi \leftrightarrow (a, b)$), the quaternions can be visualized as a four-dimensional space. These are given by things that look like $$ a + bi + cj + dk $$ where $i, j, k$ all satisfy $i^2 = j^2 = k^2 = -1$, and moreover $ij = -ji = k$. The interesting fact about the quaternions is that they are non-commutative. That is, the order in which we multiply matters!
There are also Octonions, which are even weirder, and are an 8-dimensional analogue.
Anyhow, the answer is in the end that it sort of depends. In most senses, the complex numbers are as far as you can go in a relatively natural way. But we can still look at bigger domains if we want, but we have to find other ways to build them.
Best Answer
Yes, the two sets are equivalent.
The natural numbers and the integers are constructed by very different methods, so by some arguments they are not even the same mathematical object. (For instance, one might argue that integers are the quotient set of pairs of natural numbers under an equivalence relation!)
However, at the end of the section of those constructions, authors traditionally note (or prove) that all such constructions of number systems are isomorphic to each other, and so they define completely abstract sets $\mathbb N$, $\mathbb Z$, $\mathbb Q$,and $\mathbb R$ such that we can legitimately say that $\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R$. In that sense (which is the typical sense), it would be completely legitimate to say that $\mathbb N^+=\mathbb Z^+$.