Functions

[dally165]

Hi again, I need a recap on functions (Injective, surjective, bijective) and inverse functions:

(i) f : N --> N defined by f(x) = x^3 - 1
(ii) g : Z --> Z defined by g(x) = 2x + 1
(iii) h : R --> R defined by h(x) = x(x + 3)(x - 3)

(***note that N,Z,R stands for natural #, Integer, and Real # respectively..)

(a) Which of the functions are one-to-one?


(b) Which of the functions are bijections?


(c) For those that are bijections find the inverse function.


How do I start to answer the questions? All I know is that a function is a bijection iff the function is both injective and surjective.
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Here's another one:

The function f : R --> R defined by f(x) = (3x - 1)/(x - 3) is not bijective however by suitably restricting the domain and codomain the function can be made to be bijective.

(a)State the domain and codomain that will make the function bijective.

What's a domain? codomain?

(b) Find the inverse of the bijective function.

(I can still remember a bit of inverse function.. i think.. well ill give it a try anyway)

f(x) = (3x - 1)/x - 3)
x = (3y - 1)/y - 3) "replace x with y"
x(y - 3) = 1(3y - 1)) "Cross multiplication"
xy - 3x[i/] = 3y - 1 "Will minus both sides with 3y"
xy - 3x - 3y = -1 "Will add both sides with 3x"
xy - 3y = -1 + 3x
y(x - 3) = -1 + 3x "Factor out y"
y = (-1 + 3x)/(x - 3) "Divide both sides with (x - 3)"

Is that correct??

 

[daon]

Graph some points of each function to get the feel of them. "one to one" is the same as injective, and "onto" is the same as surjective.

To show injectivity, assume f(n)=f(m) and deduce that m=n. For example, f(x)=x^5+6 as a function from Z to Z is injective, since f(x)=f(y) means that x^5+6 = y^5+6 which implies x^5=y^5. Its not exactly trivial, but with general knowledge of the integers, this implies x=y (if you'd like, show that x>y and y>x both produce contradictions).

To show surjectivity, find an element in the codomain of your function, and show the function sends an element to it. The above function is NOT surjective since otherwise we can pick the integer 1 and have 1=x^5+6 iff -5 = x^5. -5 is no 5th root of any integer. The function g(x) = x-1 is surjective however. Since if n is an integer in the codomain, g(n+1)=n.

Inverses of bijective functions are ound the same way you did in algebra, college algebra, precalc, etc.

For the second, the domain is the set your "plugging in" numbers from. The codomain is the general governing set the function sends things to. For example the floor function takes a real number and outputs an integer. Its domain is the real numbers and its codomain is the integers. There are some numbers you cannot evaluate a function at, which we exclude from the domain automatically (sometimes we leave these bad values in the domain and assume they're sent to the "empty set"). In this example, 3 is not a member of the domain, so there is no number corresponding to 3 in the codomain. Do not confuse the range with the codomain, range is a subset of the codomain. It is only equal if the function is onto.

If you are not sure if they are inverses. Take one and plug it into the other. After simplifying to should get the identity function, i.e. x. Your steps look valid to me though.

 

[dally165]

Thanks for that.. Ok let me have a try:


(i) f : N --> N defined by f(x) = x^3 - 1
(ii) g : Z --> Z defined by g(x) = 2x + 1
(iii) h : R --> R defined by h(x) = x(x + 3)(x - 3)

(***note that N,Z,R stands for natural #, Integer, and Real # respectively..)

(a) Which of the functions are one-to-one?

(i) is one to one because f(x) = f(y) and x^3 = y^3 "If it was Z instead of N then it won't be one-to-one??"
(ii) is one to one because if x=y, then x multiplied by 2 is equal to y multiplied by 2
(iii) i'm not too sure yet.. i'm guessing it is also one-to-one since if x=y works??

(b) Which of the functions are bijections?

I know that the definition for a bijective function is that when the function is both injective and surjective.. I'm still quite confused with surjection..

(c) For those that are bijections find the inverse function.

Can't work this part yet unless I have answered question b..

 

[daon]

i)is right, and would still be 1-1 if it were Z->Z. note, as just an observance, this f is not defined for 1 (if N={1,2,3,...} or 0 if N={0,1,2,...}).
ii) you proved the converse of 1-1, which is proving a function is "well defined" (meaning it is in fact a function). You need to show 2x+1=2y+1 => x=y. if f is stated to be a function, the converse is assumed true.
iii) simple rule for you (to investigate yourself and not just "say"): all ODD degree polynomials are onto when the domain and range are R. in general they are not 1-1. also, all EVEN degree polynomials are NOT 1-1 nor onto when the domain and range is R. the fact that this polynomial has more than one real root means it is not 1-1 (it is 0 three times).

If f:A->B, f is onto if and only if f "connects" or "hits" every element of B when evaluated at elements of A. Think of going to a store that has exactly ten different $1 items (but enough stock so that you may buy 15 of any distinct item). You have $15 and decide to spend it all. If you buy 9 different items, you will have 6 extra items that are duplicates to some item, right? This demonstrates a NON onto function, since you didn't buy every item at least once (one type of item remains unbought). If you do buy every item with your $15, then you have 5 duplicates, this IS an onto function... you matched your 15 $1 bills with all 10 items sold at the store.

The fact that you have duplicates at all is an example of a NON one-to-one function. Five of your $1 bills eventually went to the type of items you've already selected (sometimes called the pigeonhole property). In contrast, if you had only $10 to spend and bought all 10 items, this is an example of BOTH a 1-1 and onto function. Each of your dollar bills went to a unique item, and you had enough to buy them all.

I suppose for completeness, say you had only $9 and there are 10 items. You can still make a 1-1 correspondence buy buying 9 different items, but you can never make an onto correspondence (you're too poor ). If you decide to buy 8 different items but spend $9, this function is not 1-1 nor is it onto.

In this example, your wallet is the domain, and the store's inventory in the codomain. The Range is the set of unique items you purchase. If you buy 15 of the same item, your range consists of only that item (not 15 duplicates).

Whew.

Now a "Real" example: y=x^5 with domain and range R is both 1-1 and onto: every real number has a 5th root and a 5th power, and it is unique.

y=x^2 is not 1-1 nor onto: -2 and 2 goto ("buy") the same number ("item") in R ("the store") under the function ("by your choice."), so it is not 1-1. Also, no x value ("none of your dollars") will make x^2 equal to -1 ("buy this item"), so it is not onto.

 

[dally165]

Haha, thanks so much for that!! Very good illustration.. but allow me to take this slowly.. let me use the first example:

(i) f : N --> N defined by f(x) = x^3 - 1

So it says N ("Before") --> N ("After")
f(x) = x^3 - 1

Let me pick an N.. say 5.. so 5 is the N ("Before") and when it goes through the function f(x) = x^3 - 1, the result is 124 which is the N ("After")..
It is 1-1 because as we move up the N ("Before") say 6 it produces a unique N ("After") then 215? Is that what it is??
It is a surjection because each of the N ("Before") went to a unique N ("After")?

Can you give me another example of a 1-1 function but not a surjection?

I'm sort of getting your illustration but may be applied with a wrong thinking.. i'm in the process of getting everything straight..

 

[daon]

Your provided an example that f sends 5 and 6 to different elements. If your domain were just {5,6} that would be a complete proof, but your domain is the set of all positive integers (I'm assuming the definition that N starts with 1). We need to show that if m and n are sent to the same element, then m and n must be identically the same. So say f sends m and n (with both bigger than or equal to 2) to m' and n' respectively.

We are assuming these numbers m' and n' are the same, so m'=n'. The function's rule makes m'=m^3-1 and n'=n^3-1, since m'=n', we have n^3-1=m^3-1 => m^3=n^3. We 3 possibilities: m>n, m<n or m=n. One and only one of these HAS to be true for ANY real numbers m,n. If we show two are wrong then the remaining HAS to be true. This is "obvious" but takes some work to prove if it isn't available to you.

If it were true that m>n then since m,n > 0 we have the property that m^k > n^k for any power k, but we're only considering k=3 (I'm assuming you can use this, you may need to prove it). This contradicts that m^3=n^3, so it is wrong. In a similar manner we get that m<n produces that m^3 < n^3. Therefore the only possibility is that m=n.

So I have shown that if m,n>=2 then f(m)=f(n) implies m=n. In other words, images of every value m>=2 under this function are unique to that m. If m=1 we have an issue, as f(1) isn't defined.

As far as surjections, they don't need any kind of uniqueness. The function sin(x) is onto when the codomain is [-1,1], but it hits every value infinitely many times. You just need to make sure every element of the codomain has something sent to it.

Maybe another analogy? This is the standard one: if you have a classroom with 30 desks and you force 100 people into the room to sit down, every seat has at least one person in it. If you can "fill the desks" the function is onto. It is 1-1 if you can seat everyone one to a desk, even if every desk isn't occupied. If you can't fill all the seats, it is not onto. If every desk is occupied, and only one in each desk, it is 1-1 and onto...