How Do Cardinal Number Exponents Distribute Over Multiplication?

[saadsarfraz]

Homework Statement

prove that (a x b)^{}c = (a^{}c x b^{}c where a,b,c are any cardinal numbers

Homework Equations

The Attempt at a Solution

i know that they should first be interpreted as sets A,B,C but what functions should I use.

 

[tiny-tim]

Hi saadsarfraz!

(use the X² tag just above the Reply box, instead of tex )

prove that (a x b)^{}c = (a^{}c x b^{}c where a,b,c are any cardinal numbers

A^C is the set of functions from C to A …

so pick a typical function on one side of the equation and show how to define a corresponding function on the other side

 

[saadsarfraz]

Hi and thanks but I am still a bit confused so is this how i shld do it:

A^c is defined as f: C --> A and I also define B^c as g: C --> B and h can be defined as h: C --> A X B, and since assuming f and g are bijections h too is a bijection. Am I in the right direction?

 

[tiny-tim]

A^c is defined as f: C --> A and I also define B^c as g: C --> B and h can be defined as h: C --> A X B, and since assuming f and g are bijections h too is a bijection. Am I in the right direction?

No, not at all …

A^C is the set of all functions from C to A, not one function

 

[saadsarfraz]

i still don't know how to do it, can you please help me in this.

 

[tiny-tim]

i still don't know how to do it, can you please help me in this.

Start:

Let f:A → C be a member of A^C and g:B → C be a member of B^C …​

and then construct a member h:AxB → C of (AxB)^C using f and g

 

[saadsarfraz]

is h going to be like this h(a,b) = (f(a),g(b) next to show that this is an injection?

 

[tiny-tim]

is h going to be like this h(a,b) = (f(a),g(b) next to show that this is an injection?

Hi saadsarfraz!

Yes, that's exactly right!

('cept you missed out a bracket! )

ok, now the other way round …

starting with an h, how do you define an f and g?

 

[saadsarfraz]

I don't know how to do define an f and g starting with an h?

 

[tiny-tim]

I don't know how to do define an f and g starting with an h?

Hint: if h:C→ AxB is a member of (AxB)^C, then define the projections h_A:C→ B and h_B:C→ A

 

[saadsarfraz]

h_a going to be (B^c) and h_b(c) = (A^c)

 

[tiny-tim]

h_a going to be (B^c) and h_b(c) = (A^c)

mmm … i suspect you've got it …

but what you've actually written makes no sense​