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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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    $\begingroup$ Let me tell you more surprising thing. There are as many rational numbers as natural numbers. $\endgroup$
    – user45099
    Mar 26, 2013 at 10:10

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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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    $\begingroup$ I understand this is a valid proof but it still doesn't satisfy my intuition. I can accept that, considering the definition of cardinality, the cardinality of the set of natural numbers is the same as that of the set of even numbers. But still, in SOME sense, it must be true that the set of natural numbers is larger than that of even numbers. In fact, in some sense, it must be true that the set of natural numbers is twice as large as the set of even numbers. Is there any other concept in mathematics (an alternative to cardinality) that describes this rather universal intuition? $\endgroup$
    – Ariel
    Mar 14, 2018 at 10:49
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    $\begingroup$ @Ariel Here's something that you might find satisfying: there are multiple notions of the "density" of a set of numbers (see here or here, for example). Let $A\subseteq\Bbb{N}$ be the set of positive even integers. Then the natural density $d(A)$ of $A$ exists and is equal to $1/2,$ as you would intuitively expect. $\endgroup$
    – Stahl
    Jul 20, 2020 at 23:49
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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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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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  • $\begingroup$ Why @bryansis2010 ? Can I know the reason??? $\endgroup$ Mar 26, 2013 at 10:09
  • $\begingroup$ Ok..... @bryansis2010 $\endgroup$ Mar 26, 2013 at 10:13
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    $\begingroup$ Why do people learn probability, stats and calculus before being introduced to the concepts of cardinality and bijections...? $\endgroup$
    – Eckhard
    Mar 26, 2013 at 11:41
  • $\begingroup$ @Eckhard you might wish to phrase that as a question and ask the community here. (: $\endgroup$ Mar 26, 2013 at 12:42
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Let's stick strictly to non-negatives to avoid confusion. Here's why the concept of evens and wholes being in 1-to-1 correspondence baffles many:

We like to think intuitively that if I choose a non-negative at random, it will have a 50% chance of being even (which would imply only 50% of non-negatives are even!)

Surely, this logic is flawless right....? wrong!

The nature of the problem is the assumption one can choose a non-negative at random. I'll try to show you why this is quite a bizarre assumption and how the falsehood of this assumption creates a fatal wound in the logic above. Imagine I have a perfectly random selector that spits out a random single-digit number. If I make the assumption above, then I can use it to design a random non-negative as follows:

First, let it choose the one's place digit. e.g., "..._ _ _ 7". Then let it choose the next digit. e.g., "... _ _ 4 7". We don't immediately see the problem until we note our process can only end if at some point we randomly generate infinitely many 0's (because otherwise our number is approaching infinity).

"That's okay! We can stop at the first digit; It's odd so we KNOW our number is an odd number. This proves the above logic is correct, right?" wrong again!

Obviously if a number ends in 7, it's odd. For instance, "267" is definitely odd. But is "banana7" an even or odd number? That's an absurd question of course since "banana7" is not a number even though it ends in 7. Similarly, even though we know the process above generates something that ends in 7, there's no reason to believe we're generating a number at all. In fact, if we were generating a number x, then it would obviously have a well-defined nth and final digit (for instance 001257 has 1 as its 4th and final digit). But again, for this to happen we'd need to get lucky enough for every single digit after number n to be 0 and that's ridiculous to hope for.

This is an old question, I know, but I couldn't resist commenting when I saw that no one gave an intuitive answer to this question. Don't get me wrong, all the answers I see are amazingly clear. But I am hoping my answer will help those uncomfortable with the technical language used above.

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For the first question. let me try to give the intuition which I thought of to satisfy the same issue I had with digesting it. Try to do counting without using natural numbers. Consider the subset of natural numbers between 1 and 100. Can you count it? Yes! you can. Do not assign any natural number to it.

Can you count the number of even numbers in that subset? Yes! you can. Without assigning a natural number to it you would know that at an abstract level both the counts are same.

In the same fashion, you would also digest the fact that the cardinality of set of rational numbers is the same as that of natural numbers.

Think of counting without the handicap of natural numbers.

The second question has been answered above.

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