Question

Is there a bijection between the natural numbers and the integers? If there is one, can you please give me an example? I'm drawing a blank.

Answer 1

hmm not sure what you mean

all integers are just a subset of the natural set

so to that extent, there's a bijection, or one-to-one, if you apply restrictions to the real number set to only non-decimals

Answer 2

is there an example that takes {1,2,3,4,......} to the integers {....-3,-2,-1,0,1,2,3......} I can't seem to find one that is working.

Answer 3

hmm, not sure... what's being asked

Answer 4

The question says: "Is there a bijection between the natural numbers N and the integers Z? If so, provide such a bijection. If not, explain carefully why not.

Answer 5

http://www.mathsisfun.com/definitions/natural-number.html <----

I misread you earliers... kinda read real numbers

Answer 6

"is there a bijection between the natural numbers N and the integers Z"

meaning

the natural numbers, which are really a subset of integers, pretty much just from 0 onwards, so all positive integers

can you match them up in a one-to-one relation with the integers set?

well, since the natural numbers are a set of all integers
and the natural numbers are a subset of all integers, just from 0 onwards

then they'd intersect, or biject, from 0 onwards

Answer 7

ahemm.. lemme reword that a bit

rather

"is there a bijection between the natural numbers N and the integers Z"

meaning

the natural numbers, which are really a subset of integers, pretty much just from 0 onwards, so all positive integers

can you match them up in a one-to-one relation with the integers set?

well, since the \(\bf integer\) numbers is a set of all integers
and the \(\bf natural\) numbers is a subset of all integers, just from 0 onwards

then they'd intersect, or biject, from 0 onwards

Answer 8

so there is no bijection that can take the natural numbers to the negative integers but there is one that can take it to the positive integers?

Answer 9

yeap

Answer 10

\(\large \begin{array}{ccllll}
\mathbb{N}&\mathbb{Z}\\
\hline\\
&-4\\
&-3\\
&-2\\
&-1\\
0&0\\
1&1\\
2&2\\
3&3\\
4&4\\
+\infty &+\infty

\end{array}\)

Answer 11

thank you!

Answer 12

yw

Answer 13

One might think Z should be bigger than N but they actually have the same cardinality.

The key to realising this is to write it on a clever form:
N: 0 1 2 3 4 5 6 7 8
Z: 0 1 -1 2 -2 3 -3 4 -4

Looking at the relationship between the two rows we can construct the following fuction:
\[\phi(n) =\]\[\frac{ -n }{ 2 }\]if n is even.
\[\frac{ n + 1 }{ 2 }\]if n is odd.

Inserting a \[n \in N\]will always result in a single vaild \[\phi \in Z\]thus there is N for every Z.

Also, they are both countably Infinite.

Answer 14

that will only give negative numbers when n is even

Answer 15

wait i think i see it

Answer 16

what about zero? zero isn't a natural number but it is an integer so I need something to give zero

Answer 17

That's not entirely true. There are different opinions regarding this as N can be defined as the positive integears (0, 1, 2, 3, 4, ..) or the non negative integears (0, 1, 2, 3, 4, ..).

The first definition is mostly used in number theorey while the later preferred in set theory and computer science.

Answer 18

Regardless of what the definition is though, you can still construct a bijection. Just take\[\phi(n)=\begin{cases}
\frac{n-1}{2},\,n\text{ odd}\\
-\frac{n}{2},\,n\text{ even}
\end{cases}\]instead (notice the n-1/2 instead of n+1/2).

Answer 19

oh gosh thank so much! That helps alot!!!!