Pure And Applied Mathematics.(By Tarun Gehlot )
Mathematics is both an art and a science, and pure mathematics lies at its heart. Pure mathematics explores the boundary of mathematics and pure reason. It has been described as "that part of mathematical activity that is done without explicit or immediate consideration of direct application". i am trying to write some different in math which can create new thoughts and skills in my readers minds and they can enjoy THE MAGIC OF MATHS
Friday, 9 May 2014
Monday, 21 April 2014
DIFFERENCE BETWEEN RELATIONS AND FUNCTIONS
There are different ways of looking at functions. We will consider a few. But first, we need to discuss some terminology.
A "relation" is just a relationship between sets of information. Think of all the people in one of your classes, and think of their heights. The pairing of names and heights is a relation. In relations and functions, the pairs of names and heights are "ordered", which means one comes first and the other comes second. To put it another way, we could set up this pairing so that either you give me a name, and then I give you that person's height, or else you give me a height, and I give you the names of all the people who are that tall. The set of all the starting points is called "the domain" and the set of all the ending points is called "the range." The domain is what you start with; the range is what you end up with. The domain is the x's; the range is the y's. (I'll explain more on the subject of determining domains and ranges
A function is a "wellbehaved" relation. Just as with members of your own family, some members of the family of pairing relationships are better behaved than other. (Warning: This means that, while all functions are relations, since they pair information, not all relations are functions. Functions are a subclassification of relations.) When we say that a function is "a wellbehaved relation", we mean that, given a starting point, we know exactly where to go; given an x, we get only and exactly one y.
Let's return to our relation of your classmates and their heights, and let's suppose that the domain is the set of everybody's heights. Let's suppose that there's a pizzadelivery guy waiting in the hallway. And all the delivery guy knows is that the pizza is for the student in your classroom who is fivefootfive. Now let the guy in. Who does he go to? What if nobody is fivefootfive? What if there are six people in the room that are fivefive? Do they all have to pay? What if you are fivefootfive? And what if you're out of cash? And allergic to anchovies? Are you still on the hook? Ack! What a mess!
The relation "height indicates name" is not wellbehaved. It is not a function. Given the relationship (x, y) = (fivefootfive person, name), there might be six different possibilities for y = "name". For a relation to be a function, there must be only and exactly one y that corresponds to a given x. Here are some pictures of this:
This is a function. You can tell by tracing from each x to each y. There is only one y for each x; there is only one arrow coming from each x.  
Ha! Bet I fooled some of you on this one! This is a function! There is only one arrow coming from each x; there is only one y for each x. It just so happens that it's always the same y for each x, but it is only that one y. So this is a function; it's just an extremely boring function!  
This one is not a function: there are two arrows coming from the number 1; the number 1 is associated with twodifferent range elements. So this is a relation, but it is not a function.  
Okay, this one's a trick question. Each element of the domain that has a pair in the range is nicely wellbehaved. But what about that 16? It is in the domain, but it has no range element that corresponds to it! This won't work! So then this is not a function. Heck, it ain't even a relation! 
"Vertical Line Test"
Looking at this function stuff graphically, what if we had the relation that consists of a set containing just two points: {(2, 3), (2, –2)}? We already know that this is not a function, since x = 2 goes to each ofy = 3 and y = –2. Copyright © Elizabeth Stapel 19992011 All Rights Reserved
This characteristic of nonfunctions was noticed by Idon'tknowwho, and was codified in "The Vertical Line Test": Given the graph of a relation, if you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function. Here are a couple examples:

The domain is all the values that x is allowed to take on. The only problem I have with this function is that I need to be careful not to divide by zero. So the only values that x can not take on are those which would cause division by zero. So I'll set the denominator equal to zero and solve; my domain will be everything else.
 x^{2} – x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 or x = –1Then the domain is "all x not equal to –1 or 2".
The range is a bit trickier, which is why they may not ask for it. In general, though, they'll want you to graph the function and find the range from the picture. In this case: 
As I can see from my picture, the graph "covers" all yvalues (that is, the graph will go as low as I like, and will also go as high as I like). Since the graph will eventually cover all possible values ofy, then the range is "all real numbers".
 Determine the domain and range of the given function:The domain is all values that x can take on. The only problem I have with this function is that I cannot have a negative inside the square root. So I'll set the insides greaterthanorequalto zero, and solve. The result will be my domain:–2x + 3 > 0
–2x > –3
2x < 3
x < 3/2 = 1.5Then the domain is "all x < 3/2".
The graph starts at y = 0 and goes down from there. While the graph goes down very slowly, I know that, eventually, I can go as low as I like (by picking an x that is sufficiently big). Also, from my experience with graphing, I know that the graph will never start coming back up. Then the range is "y < 0".
 Determine the domain and range of the given function:y = –x^{4} + 4This is just a gardenvariety polynomial. There are no denominators (so no divisionbyzero problems) and no radicals (so no squarerootofanegative problems). There are no problems with a polynomial. There are no values that I can't plug in for x. When I have a polynomial, the answer is always that the domain is "all x".
The range will vary from polynomial to polynomial, and they probably won't even ask, but when they do, I look at the picture:

The graph goes only as high as y = 4, but it will go as low as I like. Then:
The range is "all y < 4"
Wednesday, 19 March 2014
What is an inverse function?
The inverse function for f( x), labeled f^{−1}( x) (which is read “ f inverse of x”), contains the same domain and range elements as the original function, f( x). However, the sets are switched. In other words, the domain of f( x) is the range of f ^{−1}( x), and vice versa. In fact, for every ordered pair ( a, b) belonging to f( x), there is a corresponding ordered pair ( b, a) that belongs to f ^{−1}( x). For example, consider this function, g:
The inverse function is the set of all ordered pairs reversed:
Only one‐to‐one functions possess inverse functions. Because these functions have range elements that correspond to only one domain element each, there's no danger that their inverses will not be functions. The horizontal line test is a quick way to determine whether a graph is that of a one‐to‐one function. It works just like the vertical line test: If an arbitrary horizontal line can be drawn across the graph of f( x) and it intersects f in more than one place, then f cannot be a one‐to‐one function.
Inverse functions have the unique property that, when composed with their original functions, both functions cancel out. Mathematically, this means that
Graphs of inverse functions
Since functions and inverse functions contain the same numbers in their ordered pair, just in reverse order, their graphs will be reflections of one another across the line y = x, as shown in Figure 1.
Figure 1 Inverse functions are symmetric about the line y = x.
Finding inverse functions
To find the inverse function for a one‐to‐one function, follow these steps:
1. Rewrite the function using y instead of f( x). 2. Switch the x and y variables; leave everything else alone. 3. Solve the new equation for y. 4. Replace the y with f ^{−1}( x). 5. Make sure that your resulting inverse function is one‐to‐one. If it isn't, restrict the domain to pass the horizontal line test.
Example 1: If , find f ^{−1}( x).
Follow the five steps previously listed, beginning with rewriting f( x) as y:
Note the restriction x ≥ 0 for f ^{−1}( x). Without this restriction, f ^{−1}( x) would not pass the horizontal line test. It obviously must be one‐to‐one, since it must possess an inverse of f( x). You should use that portion of the graph because it is the reflection of f( x) across the line y = x, unlike the portion on x < 0.
Equal Sets and Equivalent Sets
Consider two sets:
A = {−9, −3, 0, 5, 12}
B = {−2, 1, 2, 4, 7}
Did you notice any relation between the sets A and B?
Let us see.
We have,
A = {−9, −3, 0, 5, 12}
B = {−2, 1, 2, 4, 7}
Therefore, we have n (A) = 5 and n (B) = 5
Observe that both the sets A and B have same number of elements. Therefore, in this case, we say that the sets A and B are equivalent sets and it can be defined as:
Two finite sets are called equivalent, if they have the same number of elements.

Thus, two finite sets X and Y are equivalent, if n (X) = n (Y). We write it as X ↔ Y (read as “X is equivalent to Y”)
Now, consider the two sets:
X = {all letters in the word STONE}
Y = {all letters in the word NOTES}
Did you notice any relation between the sets X and Y?
Let us see.
We have,
X = {S, T, O, N, E} and Y = {N, O, T, E, S}
Observe that both the sets X and Y have same elements. Therefore, in this case, we say that the sets X andY are equal sets.
Two sets are called equal, if they have same elements.

When two sets X and Y are equal, we denote it as X = Y; and if they are not equal, then we write it as X ≠Y
Also, note that n (X) = 5 and n (Y) = 5
Therefore, we can conclude that:
If A and B are finite sets and A = B, then n (A) = n (B) i.e., A and B are equivalent. However, the converse of the above statement may not be true.
For example, if A = {2, 4, 6} and B = {1, 3, 5}, then n (A) = n (B) = 3; however, A ≠ B
Let us now look at some more examples to understand the above discussed concepts better.
Example 1:
Which of the following sets are equal?
(a) X = {x: x is a letter in the word REFRESH},
Y = {A letter in the word FRESHER}
(b) X = {4}, Y = {x: x∈ N, x − 4 = 0}
(d) X = {x/x is a vowel letter in the word WEIGHT},
Y = {x/x is a vowel letter in the word HEIGHT}
(c) X = {x: x∈ N, 0 < x < 4}, Y = {x: x∈ W}
Solution:
(a) X = {R, E, F, S, H}, Y = {F, R, E, S, H}
∴ X and Y are equal sets i.e., X = Y
(b) X = {4}, Y = {4}
∴ X and Y are equal sets i.e., X = Y
(c) X = {E, 1}, Y = {E, I}
∴ X and Y are equal sets i.e., X = Y
(d) X = {1, 2, 3}, Y = {0, 1, 2, 3, 4, 5 …}
∴ X and Y are not equal sets i.e., X ≠ Y
Example 2:
Which of the following sets are equal?
(a) X = {x: x is a vowel in the word MATRIX}
Y = {A vowel in the word SHIVANI}
(b) X = {5, 0, 1, 0, 2, 1}, Y = {3, 2, 8, 3, 3, 2}
Solution:
(a) X = {A, I}, Y = {A, I}
∴ n (X) = 2 and n (Y) = 2
This means n (X) = n (Y)
Therefore, X and Y are equivalent sets.
(b) X = {5, 0, 1, 0, 2, 1} = {0, 1, 2, 5}, Y = {3, 2, 8, 3, 3, 2} = {2, 3, 8}
∴ n (X) = 4 and n (Y) = 3
This means n (X) ≠ n (Y)
Therefore, X and Y are not equivalent sets.
Complement of a set
Let us consider a set X as
X = {2, 3, 6, 8}
Let us consider its universal set Î¾ as
Î¾ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Can we find the set of elements in set Î¾, which are not in X?
We can find this by taking the difference of X from Î¾ as
Î¾ − X = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 6, 8} = {0, 1, 4, 5, 7, 9}
This set (consisting of all the elements Î¾, which do not belong to X) is known as the complement of set Xand we denote it by or.
Let X be any set and Î¾ be its universal set. The complement of set X is the set consisting of all the elements of Î¾, which do not belong to X. It is denoted by X′ or X^{c} (read as complement of set X).
Thus, X′ = {xx∈Î¾ and x∉X} or X′ = Î¾−X

For the above sets X and Î¾, we may observe that n (X) = 4, n (Î¾) = 10, and n () = 6.
Can we find any relation among them?
We observe that:
n() = n (Î¾) − n (X) 
This relation holds true for a set, its complement, and a universal set.
The other properties of a set and its complement are as follows.
(a)
(b)
(c)
(d)
(e)
(f) If X ⊆ Y then
Apart from these properties, there are two more properties for two sets A and B. They are:
(a)
(b)
These are also known as De Morgan’s laws.
Let us prove the first one.
Let A = {1, 2, 3}, B = {2, 3, 4}, and Î¾ = {1, 2, 3, 4, 5, 6}
Now, A ∩ B = {2, 3}
Therefore, (A ∩ B)′ = {1, 4, 5, 6}
Also, A′ = {4, 5, 6} and B′ = {1, 5, 6}
∴ A ′ ∪ B′ = {1, 4, 5, 6}
Clearly, we have
Similarly, we can prove the second one.
Now, how will we represent the complement of a set A with the help of a Venn diagram?
We know that if A is a set and Î¾is a universal set for the set A, then the complement of the set A is A^{c} = Î¾− A.
If we represent the sets Î¾and A by a Venn diagram, then we can easily represent A^{c} on it.
For this, we represent the set A by using a circle and Î¾ by using a rectangle (or a square which is bigger and encloses the circle). Now, the portion outside the set A, but inside the set Î¾, represents the set A^{c}. This can be shown as follows:
Let us look at some examples in order to understand these concepts better.
Example 1:
If A and B are two sets and Î¾ is their universal set such that , and
n (B) = 6, then how many elements are there in the complement of set B?
Solution:
We know that,
We also know that,
Therefore, the complement of set B contains 2 elements.
Example 2:
If A = {x, 1, 2, 3, y}, B = {2, 4, 5, y}, and Î¾= {x, y, z, 1, 2, 3, 4, 5, 6}, then show that
Solution:
Now, A^{c} = Î¾ − A = {x, y, z, 1, 2, 3, 4, 5, 6} − {x, 1, 2, 3, y}= {z, 4, 5, 6}
B^{c }= Î¾ − B = {x, y, z, 1, 2, 3, 4, 5, 6} − {2, 4, 5, y} = {x, y, 1, 3, 6}
A ∪ B = {x, y, 1, 2, 3, 4, 5}
∴(A ∪ B)^{c} = {z, 6}
Now, A^{c} ∩ B^{c} = {z, 6}
Clearly, (A ∪ B)^{c} = A^{c} ∩ B^{c}
Example 3:
Find the following sets from the adjoining Venndiagram.
(i) (A ∩ B)^{c}
(ii) A^{c}
(iii) (A ∪ C)^{c}
(iv) Î¾
Solution:
From the given Venndiagram, we find that
(i) (A ∩ B)^{c }= {1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 19}
(ii) A^{c }= {3, 4, 6, 7, 9, 11, 12, 13, 14, 15}
(iii) (A ∪ C)^{c }= {3, 4, 7, 9, 11, 13}
(iv) Î¾ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 19}
Example 4:
Taking the set of first ten natural numbers as the universal set, find the set
(B− A)′ ∩ B′, where A = {1, 2, 4, 9} and B = {2, 5, 7, 9, 8, 10, 1, 3}
Solution:
B − A = {5, 7, 8, 10, 3}
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(B − A) ′ = {1, 2, 4, 6, 9}
B′ = {4, 6}
∴ (B− A)′ ∩ B′ = {4, 6}
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