Welcome, I don't know why you are here, but I will tell you what is this blog about so you don't waste any time looking around . This is a math blog, as you already know. Here I will expose math topics, from simple algebra, to calculus and all in between. My idea here is to make this interactive, depending on the feedback I get, so make sure to comment on the posts if you don't understand something or you want me to write about some particular subject, because my aim is to help people, not to show off all my knowledge or to shout out "cool" stuff. I know math can be boring to most people and I intend to write of the most general subjects seen in school, highschool and college in a way that anyone can understand. Here you will find some very basic stuff written in simple words, how to solve word problems, what is math, algebra, math in games, fractions, multiplication, division, math for kids, for adults, cool math, boring math, whatever I find interesting to discuss and whatever YOU want to see here, so I'm counting on you to make this interactive, let's make that happen.

17 may. 2013

Functions

 Functions.


Ok, today we are going to talk about functions. You have to be aware that this is a HUGE topic in math and here I only want to give a simple word definition on what functions are, so we can move on to linear functions, graphing and all that stuff. So a function, a function is, in simple words, a relation between two sets. You have set A (that could be a bucket of dirty socks) and set B (that could be a bucket of gloves), a function is a relation between the elements of those sets (for example, pair the sock with the glove of the same color). Now, there is a condition that a function must satisfy. All elements of the input set (the set you apply the function to) must have a corresponding element in the output set and ONLY ONE. That is why we make a difference bewteen the input set and the output set, and they are called Domain and Codomain (or Range) respectively. So you have to specify what sets you will use, and for a given relation, you can define the function with a pair of sets and not with another pair. 
Let's see real examples of this. Say set A ={1,2,3,4}  (the numbers 1,2,3 and 4) and set B={2,4,6,8}, so we define a relation, with A as the domain and B as the Codomain and the relation is as follows  f(a) = 2a. What does this mean? That for an input "a" (a is an element of A, i.e 1,2,3 or 4)
we get the outcome by multiplying it by 2. Now, is this relation a function? Well, we have to see if it satisfies the condition, does each element of A have a corresponding element in B? let's see,  
f(1) = 2*1 = 2 so, (1,2) is the ordered pair. (an orderer pair is what the words say, two things, in a particular order) Is 2 in B? yes, does 1 have any other corresponding element in B? no, if we multiply 2*1 we will get 2 and only 2. So far we have no problem, but we can't say it is a function yet, we have to see if this holds for 2,3 and 4. And it does, you can check that f(2) =4,  f(3) = 6 and f(4) = 8. All 4,6 and 8 are in B and for none of the a's (2,3 and 4) we have more than one corresponding element in B. So f(a)= 2a is a function if we use A={1,2,3,4} as the domain and B={2,4,6,8} as the codomain. What happens if we flip the roles? what if we want B to be the domain? Same relation
f(b) = 2b  ( we use b now because B is the domain, we could have used x in both cases, the a and b were to emphatize the domain). Well, it works for 2,  f(2) = 4 and the pair (2,4) with 2 in B and 4 in A is valid pair, because 4 is in A, but none of the others, f(4) = 8, 8 is not in A (and that is enough to prove it is not a fucntion, but let's see the rest)  f(6)= 12, 12 is not i n A, and  f(8) = 16, not in A.
Som the same relation, f(x) = 2x, is a function with domain A and codomain B, but it is NOT a function with domain B and codomain A. Let's see a couple of examples.
I will use X for the domain set and x for the domain's elemets and same with Y and y with the codomain.
X={1,2,3,4} ; Y={1,2,3,4}
f(x) =2
This relation (we can't call it function yet) relates all x's with the number 1. If we were to list the
ordered pairs, we would have (1,2), (2,2), (3,2) and (4,2). Is it a function? You may say no, because all elements of X have the same corresponding element in Y, you would have fallen in my trap (muahaha) the condition is that all elements of X have one and only one correspondinf element in Y. And that happens, all of them have one and only one, who cares if all have the same, there's no rule with that, there is also no rule that says there can't be unpaired elements in the Codomain. So this is indeed a function. Another example....
I will just give the orderer pairs here, same X and Y as before. I will use the notation "(x,f(x))" for the set of all the ordered pairs given by the relation.
"(x,f(x))" = {(1,1), (2,1), (2,2),(3,3),(4,4)}
Is this relation a function?? NO!! why? because 2 (the "x 2") has two corresponding elements, so not a function.
I will leave some problems for you to work, relations, you tell me if they are function or not, and if not, why? Leave your answers in the comments.

1)  X={1,2,3,4} and Y={2,3,4,5}
f(x) = x+1

2) X={1,2,3,4} and Y={2,3,4}
f(x) = x+1

3)  X={0,2,4,8} and Y={2,3,4,5} 
"(x,f(x))" = {(0,2),(2,3),(4,4)} 

4) X={1,2,3,4} and Y={2,3,4,5}
f(x) = 1







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