Question

Let A and B any two sets. Prove: A x B ~ B x A

Answer 1

A={2,3}
B={5}
Then AxB={(2,5),(3,5)}
and BxA={(5,2),(5,3)}

Answer 2

what is the definition of ~

Answer 3

or the meaning of

Answer 4

Equuimerous, meaning there exists a function f which maps X -> Y and is 1 to 1 and onto

Answer 5

oh so we want to show |AxB|=|BxA|

Answer 6

If we assume A and B are finite sets.
And that |A|=m and |B|=n.
What is |AxB|=?

Answer 7

Im not sure what you mean. The X is referring to the Cartesian product. Sorry, this is my first question I should have used better notation and wording

Answer 8

I assumed AxB is the Cartesian product of A and B.

Answer 9

I'm asking you if the number of elements in A is m and the number of elements in B is n
then number of elements in AxB is?

Answer 10

m x n

Answer 11

yes mn
and the number of elements in BxA is?

Answer 12

nm

Answer 13

and mn=nm since multiplication is commutative
we have shown |AxB|=|BxA| for finite sets A and B

Answer 14

I think that is all we really need to do...
But it might also help to find a bijective function from AxB to BxA

Answer 15

yeah I was trying to use contradiction but I don't remember how to negate ~

Answer 16

hmm... So you wanted to assume AxB is not equivalent to BxA
and wind up with a contradiction to that...hmm...

Answer 17

I feel like a direct proof is more the route to go...

Answer 18

but...

Answer 19

that might be easier actually...
we wouldn't have to find a bijective function that way I think
like if AxB is not equivalent to BxA
then there is no way AxB and BxA can have the same amount of elements
but we already showed they did

Answer 20

therefore by contradiction AxB is not equivalent to BxA is false

Answer 21

and AxB is equivalent to BxA

Answer 22

But if we wanted to do the direct proof...
We want to suppose there is a function f:AxB->BxA
the elements in AxB are ordered pairs
just like the elements in BxA
so we want f(a, b)=( b ,a )
where a is an element of A
and b is an element of B
Test:
Say we have the sets from before
A={2,3}
B={5}
AxB={(2,5),(3,5)}
BxA={(5,2),(5,3)}
our function we defined before should give us all the elements in BxA given the elements in AxB
f((2,5))=(5,2)
f(3,5)=(5,3)
fun stuff
so if you wanted to do a direct proof
all you need to do is prove f:AxB->BxA defined by f(a,b)=(b,a) where a is in A and b is in B

Answer 23

forget to say

is one to one and onto :p

Answer 24

prove that it is one to one and onto that is

Answer 25

do you know how to do that?

Answer 26

yeah, I can probably get it from here. I have anymore Ill ask you. Big help thank you!

Answer 27

have fun
and i will be a around little while longer