Is this equality about numbers true?

Question

Integers numbers are the ($\mathbb{+N}$) $\cup$$ (\mathbb{-N}$) and the $0$, so i can prove that:

$\mathbb{Z} +(-\mathbb{N})$ = $\mathbb{N}$

$\mathbb{Z} = \mathbb{2N}$

so,

the infinite $\mathbb{Z}$ is the double of infinite $\mathbb{N}$ ?

Answer 1

There have been many questions about comparing infinities before, so here is the answer I know from those questions:

• Infinity is not a number, you cannot make add, substract, multiply or divide them.

• For sets, you cannot add, substract,... either.

Let's say I own a bookstore with infinite books. You buy half of the books from me, so I got half of the books remaining, having the incorrect equation: $\infty-\infty=\infty$. This is incorrect because the result of $\infty-\infty$ is not defined (because there can be many more counterexamples).

Also, $\infty$ is not a real number, so it cannot be directly be used in any math equations, but note that the statement "The number of solutions for the equation $x^2+1>0$ is $\infty$" is correct, but $x=\infty$ is not correct.

For your question, saying that "The number of integers are twice as many as natural numbers" is incorrect, simply because you can't compare infinities, not to mention that you can't apply comparison for sets.

Answer 2

There is some correct intuition here and it only takes minor fixing to turn it into correct mathematics.

In $\mathbb{Z} +(-\mathbb{N})=\mathbb{N}$, you're probably trying to say: if one removes negative integers from $\mathbb{Z}$, you'll get $\mathbb{N}$. A correct way to express this is: $$ \mathbb{Z}\backslash (-\mathbb{N}^+)=\mathbb{N} $$ where $\mathbb{N}\equiv\{0,1,2,3,\ldots\}$ and $\mathbb{N}^+\equiv\{1,2,3,\ldots\}$. One can define addition of sets of numbers as $A+B\equiv\{a+b:a\in A,b\in B\}$ so $\mathbb{Z} +(-\mathbb{N})$ can make sense, but in this context $\mathbb{Z} +(-\mathbb{N})$ in fact equals $\mathbb{Z}$.

In $\mathbb{Z}=2\mathbb{N}$, you're probably trying to say: there are numbers in $\mathbb{Z}$ that aren't there in $\mathbb{N}$. Specifically, $\mathbb{Z}$ is formed by putting together $\mathbb{N}$ and a "flipped copy" of $\mathbb{N}$ so, in a sense, $\mathbb{Z}$ is "twice" the set that $\mathbb{N}$ is. A correct way to express this is $$ \mathbb{Z}=\mathbb{N}\cup(-\mathbb{N}). $$ (Throughout this answer, scalar multiplication of a set is defined as $xA\equiv\{xa:a\in A\}$.)