function proof

• Jul 3rd 2010, 09:38 AM
alexandros
function proof
Challenge problem

Given, the following definition of a function from a set A to a set B
denoted as: http://www.mathhelpforum.com/math-he...93d999cc73.png:

1) http://www.mathhelpforum.com/math-he...5345ddd852.png

2) for all ,http://www.mathhelpforum.com/math-he...3381e606ef.png ,there exists a unique http://www.mathhelpforum.com/math-he...9946c65024.png ,such that http://www.mathhelpforum.com/math-he...e380321460.png

Prove that the following f:

f={(1,a),(2,b)} is a function from A to B,where :

A = {1,2} and

B = {a,b}

Moderator approved CB
• Jul 3rd 2010, 11:52 AM
undefined
Quote:

Originally Posted by alexandros
Challenge problem

Given, the following definition of a function from a set A to a set B
denoted as: http://www.mathhelpforum.com/math-he...93d999cc73.png:

1) http://www.mathhelpforum.com/math-he...5345ddd852.png

2) for all ,http://www.mathhelpforum.com/math-he...3381e606ef.png ,there exists a unique http://www.mathhelpforum.com/math-he...9946c65024.png ,such that http://www.mathhelpforum.com/math-he...e380321460.png

Prove that the following f:

f={(1,a),(2,b)} is a function from A to B,where :

A = {1,2} and

B = {a,b}

Moderator approved CB

I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple $\displaystyle (A,B,F)$ with $\displaystyle F \subseteq (A\times B)$ as you defined above.

Anyway it's easy to check that {(1,a),(2,b)}$\displaystyle \subseteq(A\times B)$={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

(Fixed wording.)
• Jul 3rd 2010, 02:59 PM
alexandros
Quote:

Originally Posted by undefined
I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple $\displaystyle (A,B,F)$ with $\displaystyle F \subseteq (A\times B)$ as you defined above.

Anyway it's easy to check that {(1,a),(2,b)}$\displaystyle \subseteq(A\times B)$={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

(Fixed wording.)

I think there is a difference between ""check" and "prove",don't you?
• Jul 3rd 2010, 03:12 PM
Plato
Quote:

Originally Posted by alexandros
I think there is a difference between ""check" and "prove",don't you?

There is no difference in any finite case.
I absolutely agree, I cannot see why this is a challenge problem.
• Jul 3rd 2010, 05:28 PM
alexandros
If you want me to reveal my proof so you can see the difference ,i can do so.
• Jul 3rd 2010, 06:22 PM
tonio
Quote:

Originally Posted by alexandros
If you want me to reveal my proof so you can see the difference ,i can do so.

Please do. Without putting any shame on you, I think that including questions like this one in the challenge problems section demerit and defile the intention of it.

Tonio
• Jul 4th 2010, 12:51 AM
alexandros
proof:

Let :$\displaystyle \vee$ denote "or" ,and

let :$\displaystyle \wedge$ denote "and"

1)Let (x,y)εf =>

$\displaystyle [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ => χ=1v x=2.

For x=1

put $\displaystyle y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B$

Also

$\displaystyle (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f$

Thus there exists y such that,$\displaystyle y\in B$ and $\displaystyle (x,y)\in f$

So much for the existence part.

Now for the uniqueness part.For that we must prove:

for all ,y, z:$\displaystyle [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]$

So let :

$\displaystyle (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]$

BUT $\displaystyle 1\neq 2$ (an assumption needed in the initial hypothesis for f to be a function),thus $\displaystyle 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$

,since x=1.

In the same way .$\displaystyle (x,z)\neq (2,b)$

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

for x= 1 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

in the same way:

for x=2 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

Therefor.:

For all ,x:$\displaystyle x\in A$ there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$ .

Thus :$\displaystyle f:A\to B$
• Jul 4th 2010, 11:54 AM
tonio
Quote:

Originally Posted by alexandros
proof:

Let :$\displaystyle \vee$ denote "or" ,and

let :$\displaystyle \wedge$ denote "and"

1)Let (x,y)εf =>

$\displaystyle [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ => χ=1v x=2.

For x=1

put $\displaystyle y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B$

Also

$\displaystyle (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f$

Thus there exists y such that,$\displaystyle y\in B$ and $\displaystyle (x,y)\in f$

So much for the existence part.

Now for the uniqueness part.For that we must prove:

for all ,y, z:$\displaystyle [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]$

So let :

$\displaystyle (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]$

BUT $\displaystyle 1\neq 2$ (an assumption needed in the initial hypothesis for f to be a function),thus $\displaystyle 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$

,since x=1.

In the same way .$\displaystyle (x,z)\neq (2,b)$

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

for x= 1 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

in the same way:

for x=2 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

Therefor.:

For all ,x:$\displaystyle x\in A$ there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$ .

Thus :$\displaystyle f:A\to B$

With all due respect the above is completely absurd and if I may add, even ridiculous: nobody not out of his mind would ever present such a cumbersome, long and ridiculously messy proof of such a basic, easy-to-check and straighforward fact in such a basic case, UNLESS he's been said to do so in order to check his/her command in this or that logic-related or definition-related matters.

Anyway, and as has already been pointed out by some other members of the forum, unless some really interesting, unexpected and/or brilliant insight is produced I cannot understand how this question has been aproved as a challenge one, and even less can I understand how the above horrible, awful proof could possibly be accepted by any teacher/lecturer or professor in any university.

Tonio
• Jul 4th 2010, 02:47 PM
mr fantastic