Discrete Math (functions) help, please?
1) aRb ↔ a ≡ b (mod n). Test for reflexivity, symmetry, and transitivity.
2) Show that g(x) = e^x is one-to-one but not onto.
3) Composition of function is associative, i.e.,
If f: A → B, g: B →C, and h: C → D,
h ⁰ (g ⁰ f) = (h ⁰ g) ⁰ f.
Show that it is not commutative.
* In number 1, reflexivity is shown by aRa; symmetry by aRb ↔ bRa; transitivity by aRb ↔ bRc ↔ aRc (something like that) but I don't know how to apply it in the given problem.
* In number 2, a function f: A→B is said to be onto iff f(A)=B. A function f: A→B is said to be one-to-one iff f(x₁)=f(x₂) implies x₁=x₂.
* In number 3, I know that composition of function is associative. Please help me prove that is is not commutative.
Please experts, help me! Thank you very much in advance! Have a good day! :)
- MathMan TGLv 79 years agoFavorite Answer
1) a R b means "a and b have the same remainder when divided by n"
all the even numbers have remainder 0 when divided by 2
and all the odd numbers have remainder 1 when divided by 2,
so 3 R 5 (for n = 2)
but not 3 R 4 since they have different remainders.
For a R b ... the three properties all hold by virtue of their holding
Reflexive: a R a since a (mod n) = a (mod n)
Symmetric: a (mod n) = b (mod n) → b (mod n) = a (mod n)
by reflexivity of equality,
and similarly for Transitive:
Let a' = a (mod n), b' = b (mod n), c' = c (mod n)
a' = b' and b' = c' → a' = c' by transitivity of equals.
e^x = e^y → x = y
(the value of x and y being ln (e^x) or ln (e^y))
but since 0 does not have a logarithm in any base, including e,
there is no x such e^x = 0
hence it is not onto.
3) One simple counter example will suffice to show non-commutativity:
Let f(x) = x^2
Let g(x) = x+1
Let x = 5
f( g(5) ) = (5+1)^2 = 36
g (f(5) ) = (5^2) + 1 = 26
f( g(x) ) = (x+1)^2 = x^2 + 2x + 1
g( f(x) ) = x^2 + 1
those are only equal for x = 0