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I just read somewhere that there are as many even numbers as natural numbers! Is there an intuitive way of coming to terms with this fact? Because honestly, it's quite baffling. Also, does the concept of odd/even apply to negative integers as well? By definition, it should. But I have never seen a negative multiple of two being used as an example of an even number.

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Let me tell you more surprising thing. There are as many rational numbers as natural numbers. –  user45099 Mar 26 '13 at 10:10

3 Answers 3

up vote 4 down vote accepted

Let’s take the questions in reverse order.

Yes, the concepts of odd and even apply to negative integers: any integer $n$ is even if and only if there is an integer $k$ such that $n=2k$. The integer $k$ can be positive, negative, or zero. Thus, $-6=2(-3)$ is even, as is $0=2\cdot0$. An integer is odd if and only if it is not even, so $-7$ is odd: there is no integer $k$ such that $-7=2k$.

The answer to the first question is also yes. First, it’s very easy to put the set of natural numbers, $\Bbb N=\{0,1,2,3,\dots\}$, into one-to-one correspondence with the set $E=\{0,2,4,6,\dots\}$ of even natural numbers: the map $\Bbb N\to E:n\mapsto 2n$ is clearly a bijection. If you want to fine a bijection between $\Bbb N$ and the set of all even integers, you have to work a little harder. Here’s a picture of part of one:

$$\begin{array}{r} 0&1&2&3&4&5&6&7&8&9&\dots\\ 0&2&-2&4&-4&6&-6&8&-8&10&\dots \end{array}$$

This can be expressed as

$$n\mapsto\begin{cases} -n,&\text{if }n\text{ is even}\\ n+1,&\text{if }n\text{ is odd}\;, \end{cases}$$

and you can check that this really is a bijection from $\Bbb N$ to the set of all even integers.

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You can have a bijection from the set of Natural numbers($\mathbb{N}$) to the set of even no.($\mathbb{E}$) by sending each natural number $n$ to $2n$.

So this implies "there are as many natural numbers as there are even no.".

Intuitively you can think of this in this way: Any positive even no. is of the form $2n,n\in N$ So if you divide this even number by $2$ you get an $unique$ natural no. And if you multiply each natural no. by 2 you get an $unique$ even number, so their "cardinalities " must be same.

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Why @bryansis2010 ? Can I know the reason??? –  Abhra Abir Kundu Mar 26 '13 at 10:09
    
Ok..... @bryansis2010 –  Abhra Abir Kundu Mar 26 '13 at 10:13
    
Why do people learn probability, stats and calculus before being introduced to the concepts of cardinality and bijections...? –  Eckhard Mar 26 '13 at 11:41
    
@Eckhard you might wish to phrase that as a question and ask the community here. (: –  bryansis2010 Mar 26 '13 at 12:42

Suppose we have two finite sets, $A$ and $B$. If we want to check if these sets are of the same size, we can do two things. The first option is to simply count the number of elements in both sets and compare these numbers. The second option is to see if we can find a relationship between $A$ and $B$ such that, under this relationship, every element in $A$ corresponds to exactly one element in $B$, and every element in $B$ corresponds to exactly one element in $A$. This kind of relationship is called a bijection (or: one-to-one correspondence).

Using the concept of a bijective relationship, we can show that the set of all positive integers and the set of all even integers are of the same 'size'. As others pointed out, you can even show that there is a bijection between the set of rational numbers and the set of positive integers. Hence, these sets are also of the same 'size'. If you want to see how this works, see here.

In my opinion, the most baffling thing is not that these sets are of the same size, but that there exist infinite sets that are not bijective to other infinite sets. For example, the set of all real numbers $\mathbb{R}$ is not bijective to $\mathbb{Q}$, the set of all rational numbers. So there exists different 'levels of infinity'.

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