Identify the following relation on ℕ as one-to-one, one-to-many, many-to-one, or many-to-many:
R = {(x,y) | x = y+1}

- anonymous

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- anonymous

@zepdrix hey buddy, feel like helping me do a little math?

- phi

if you know what number y is , can you find x ?

- anonymous

yes

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- phi

if you know x, could you find y ?

- anonymous

yes

- phi

if you can do that, then it is one-to-one.

- anonymous

okay, thank you. it wasn't put so simply in my text :X

- anonymous

@phi Calculate the following composition function using the functions f and g defined below:
f:N→N, f(x) = x2 and g:N→N, g(x) = 2x - 3
(g∘f)(75)
would you also mind showing me how to solve this kind of problem, please? does this mean that f(x) and g(x) = 75?

- phi

do you know what f(1) means ?

- anonymous

that f(x)=f(1)

- phi

yes, and do you know how to "evaluate" f(1) ? i.e. what number is that ?

- anonymous

you plug it into all the x for f(x). so, i just plug 75 into both sides?

- phi

f(x) is the "name"
\(x^2\) (I assume you mean this and not x*2)
is the "rule" or instruction
If we say f(x) we are using a "short name" for x^2
if we say f(1), that means replace x with 1 in the formula: we get 1^2 = 1*1= 1

- phi

(g∘f)(75)
is just ugly syntax that means
g( f(75) )
which is also a bit ugly.
but we tackle it in steps
for example, what is
f(75) ?

- anonymous

5625

- phi

yes, and that means
g( f(75) )= g(5625)
g(x) = 2x -3
what is g(5625)

- anonymous

11247

- phi

that is how you do it.

- anonymous

so its left to right?

- anonymous

it would have been different if i had done the other side first

- phi

you can also do this
(g∘f)(75)
first do (g∘f)(x) which means
g( f(x) )
that means replace x in the formula for g with f(x)
g(x)= 2x-3
g( f(x) )= 2* f(x) -3
but f(x) is x^2 so we get
g(f(x))= 2*x^2 -3
now we can do
(g∘f)(75) = 2*(75)^2 -3

- phi

yes, there is no guarantee we get the same answer for
(f∘g)(75)

- anonymous

you're sure about this?

- phi

yes, the process is correct. I am not sure what you mean by f(x)=x2
that might mean x^2 or x*2, but you will have to say which you meant. (x2 is not "legal" but lots of people don't know how else to write x^2 i.e. \(x^2\)

- anonymous

its ^2

- phi

normally people don't pick big numbers for x (like 75) because you get huge numbers, and the point is not to learn arithmetic, but how to compose functions.

- anonymous

Okay, that is pretty straight forward. I was just confused about what the symbol meant and which came first; or, if i was going to get two answers. I have one other things i need to clear up and that is determining weather sets are reflexive, symmetric, transitive, or asymetric.
You did a pretty great job explaining 1 to 1 earlier. I wonder if you can help me out with a couple other questions.
Symmetric: For every element x or y in the set, if (x,y) is in the
relation then (y,x) is also in the relation.
Antisymmetric: For every element x or y in the set, for every case where
(x,y) and (y,x) are both in the relation, x = y.
Reflexive: For every element x in the set, (x,x) is in the relation.
Transitive: For every set of three elements x, y, and z, if (x,y) and
(y,z) are in the relation then (x,z) is also in the relation.
^^ these are my definitions
Let S = {1,2,3}. Determine the truth value of the statement:
The following relation is symmetric on S.
R1 = {(1,3), (3,3), (3,1), (2,2), (2,3), (1,1), (1,2)}
and that is my problem.

- anonymous

Symmetric: For every element x or y in the set, if (x,y) is in the
relation then (y,x) is also in the relation. i actually do not understand what this means. if the x,y is in relation that means it's in an ordered pair, right?

- anonymous

@phi

- phi

a relation has an "in" and an "out"
you start with x (the input) , and use a formula to find y (the output)
(x,y) means we put in "x" and got out "y"
(y,x) means we put in "y" and got out "x"

- anonymous

so, this set isn't symmetric because you can put in x = 1 y = 2 but you cant put 2-x and y = 1

- phi

yes, exactly. it has (1,2) but not (2,1), so not symmetric

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