What is Inverse Variation

In the given two variables if the value of one variable increases and another one decreases   that is if the values change in opposite manner then it’s said to be inverse variation.
For example if the speed increases the travelling time decreases and if speed decreases the travelling time increases

Inverse variations  occurs  when two variables such as XY are always equation to some constant K ie, one value  decreases and the other value increases

No of workers Days to complete the  work
        15               15
        10               18
         5               20

As the   number of worker decreases the days to complete the work increases.  Here inverse variations occur.

Inverse Variation Formula

Inverse-variation between two variables say  y and x is given by the formula    y  =  k/x
Here the variable K is known as the constant of proportionality.  Y varies inversely with x.
Y varies inversely with x, if there is a nonzero constant k such that xy = k or ,  y  = k/x, where k ≠ 0.
In inverse-variation its seen that , when the value of one  variable increases the value of the other value  decreases in proportion so that the product remains the same always.

Inverse Variation Word Problems

While solving inverse-variation word problems we use the formula ‘y = k/x   or any other variables relevant to the problem. Find the value of ‘k’ which is the constant of proportionality.
Plug in the value of k in the formula y=k/x .  Use the information from the given problem and solve for the variable.  An example problem is solved below.

If a car travels at the rate of 20 miles per hour it takes 1.5 hours to reach its destination.  If the same car travels at the rate of  50 miles per hour  what will be the time taken to reach its destination.

Use  the formula     t  =  k/s  where  t = 1.5 and s = 20 .  Plug in the values of t and s and find k.
         
                                   1.5  = k/20    (multiply by 20 both sides)
                                   K  = 1.5 x 20
                                   K =  30

Substitute the value of ‘k in the  equation   t = k/s
Here   k = 30  and  t  =  50
                                    t = 30/50

                                    t =  0.6 hours
The car takes 0.6 hours if it travels at the speed of 50 miles per hour.    Here it is found that as the speed of the car increases the time taken reduces.  So   there is inverse-variation here.

Dot product

There are two types of multiplication of two vectors. That is, as two types of products. One of them gives the result as a vector quantity and the other results as a scalar quantity. They are termed as cross product and dot product respectively.  The names are given so because in the first case a ‘x’ is used to indicate the multiplication and in the second case a ‘•’ indicates the same operation. Because of the fact that a dot product results in scalar quantity, the dot-product is also called as scalar dot product.

If P and Q are two vectors, then the dot-product two vectors is denoted as P • Q and its magnitude is given by lPl*lQlcos θ, where θ is the angle between the vectors P and Q. Unlike vector products, scalar products of vectors are commutative.

A typical example of a scalar product is the calculation of work done W by a force F while displacing the object by a displacement D. As per physics concept, W = lFl*lDlcos θ, where θ is the angle between the force and the direction of displacement. Although, F and D are vector quantities, the work done W is a scalar quantity. The work done by a force has no direction. .

From the definition, P • Q = lPl*lQlcos θ, let us analyze when the dot product is zero. It is obvious that it happens when P or Q or both are 0. But there is one more possibility. The scalar product becomes 0 when the angle between the vectors is 90o, even though P and Q may have finite values, since cos 90o = 0. In other words, the scalar product of two vectors acting at right angle is 0. We can illustrate this fact with a practical example. We all know that the weight W of an object placed on a smooth horizontal surface is a vector acting vertically. Suppose the object is moved horizontally by a displacement D, the vectors W and D act at right angle. Since the scalar product is 0, the work done W is 0. That is, no work is needed to be done in moving an object at a horizontal level subject to the assumption that there is no friction between the object and the horizontal surface. That is why we find that only a little amount of effort is needed to push even a heavy object horizontally. It is not zero as per theory, because the assumption of ‘no friction’ is practically impossible and the little effort only accounts for work done against the frictional force.

Mathematical Induction Definition

It is a process that is used to trueness of a given statement for all positive integers. This method includes proving the first statement true firstly from the series of infinites statement and then it is proved that the any one of the other statements from that series is also true.

The Principle of Mathematical Induction includes following steps to prove that a statement is true for all natural numbers:

1)Basis step: the statement is proved true for the least natural number which is normally 0 or 1.
2)Inductive step: In this step it is assumed that the statement is true for some natural number p that lies between least possible value of n to any natural number n, and then it is proved that the statement is true for the next natural number (p+1) also.
The value of n for basis step is decided from 0 and 1 on the basis of given question or statement.

Proof of Mathematical Induction
Let statement S(n) is false for some values of n. Let no be the least value of n such that S(no) is false. no cannot be 0, because S(0) is true. Therefore no should be 1+n1. As n1
Let us take some Examples of Mathematical Induction to solve some problems:

Mathematical Induction Proofs Examples

Q.1) Show that the sum of first natural numbers from 0 to n is given by n(n+1)/2.
Sol.) The above statement can be written mathematically as:
0+1+2…..+n = n(n+1)/2

Let us use mathematical indction process now:
Basis step: in this step we will put least possible value of n in above equation, which is 0 for this case.
S(0) =>
 0 = 0(0+1)/2
In above equation LHS is 0 and RHS also solves to 0. Hence basis step hold true and thus statement is true for n=0.

Inductive step: let us assume that the equation is true for a number p that lies between 0 and n. Now we will show that it is true for p+1 also.
S(p+1)=>
0+1+2…..+p + (p+1)= ((p+1)((p+1)+1))/2
LHS can also be written as
(p(p+1))/2 + (p+1) , as we assumed that the statement is true for p natural number.
(p(p+1))/2 + (p+1) = (p(p+1)+ 2(p+1))/2
= (p^2+p+2p+2)/2
= ((p+1)(p+2))/2  
=((p+1)((p+1)+1))/2 = RHS.
Therefore S(p+1) is true. Hence from basis and inductive steps we come to the conclusion that S(n) is true for all natural numbers.

LCM and its methods

LCM of two numbers is the smallest number (non zero) that is multiple of both.

When we add or subtract any fraction, we make use of the least common denominator and least common denominator is nothing but the least common multiple of more than two numbers.

LCM - least Common Multiplies actually the least possible common multiple of two or more than two numbers. We can find the least common multiple by using two methods: -

Methods for finding LCM of numbers:-
1. LCM Using Prime Factorization– In this method we find the prime multiples of all the numbers for which we need to find the Least common multiple.
We find the common factors among them and the uncommon factors. Least Common Multiple is product of common factors and product of uncommon factors.

LCM Examples- Suppose we need to find the least common multiple of 15 and 25.
We see that 15 = 3 times 5 and 25 is 5 times 5.
Now we see common factor is 5 and uncommon factors are 3 and 5.
So the least common multiple will be 3 times 5 times 5 which equal 75. Hence the least common multiple of 15 and 25 is 75.

2. Common division method- The other method of finding the least common multiple is the common division method in which we arrange the numbers together separated by commas.
We start dividing with the smallest prime number and go on dividing, till none of the numbers can be divided any further.

For example:- If we need to find LCM of 20 and 30 by common division method, we first divide both of them by 2, we get 10, 15. Then we divide again by 2, we get 5, 15. Now we divide by 3, we get 5, 5 and lastly by 5, we get 1, 1.
Now we multiply all the prime numbers by which we divided. They are 2, 2, 3, 5 which is 2 times 2 times 3 times 5 and that gives 60. Hence the least common multiple of 20 and 30 is 60.

We can try different problems for LCM Practice.
We can find the least common multiple of more than two numbers also.
This is usually helpful when we are adding and subtracting the fractions.

Rounding Decimals

Teaching rounding decimals is very easy it helps us in day to day calculations as well. So let us learn how to round the decimals. A decimals rounding is nothing but using decimals and making them a whole round figure. Which makes math’s more easy and understandable. Let us use the example to understand rounding decimals. As we have already studied rounding decimals lesson in our school, for round to decimals let us Say, how to round 66.99 to the nearest tenth? So the number is 66.99 and we have to round to nearest “Tenth”. We will be doing by using four simple steps. So step 1.  Is to identify the digit to be rounded. In this case it is 66.99, and we are asked to round the nearest tenth. The tenth digit is 66.9 the very first 9 to the decimal, because it is one tenth of the whole. Step 2. Is look at the digit at the immediate right of the one to be rounded that it another 9 in this case. Step 3. Determine the digit if it is greater than 5 or less than 5. If greater than 5, we add 1 to the digit being rounded, and if it is less than 5, we leave it unchanged. The digit is greater than 9 here, and 9 is greater than 5 so we will add 1 to it at the tenth place. So 9+1 gives 10, 0 goes after the decimal and 1 gets carried over to 6 now add 66+1 is 67.09. Step 4. We replace all digits to the right one being rounded with zeros. Hence it will become 67.00 or 67 after rounding of decimals.

Example, we buy a bag full of taco chips costs $ 0.4582 per ounce. How much is this to the nearest cent? How do you Round a Decimal? Lets us solve it. Our question says a bag of taco chips is 0.4582 dollars; we need to round to the nearest cent. So a cent is 1/100 of a dollar. Step 1. This is to identify the digit to be rounded. It is the cent digit which is 5( in 0.4582). In case of 0.4572 the zero to be rounded. Here the digit to be rounded is 5. Step 2. Look at the digit to the right of it. Our immediate right of 5 digit is 8. Step 3. We look at digit 8 and determine that if it is less than 5 or greater than 5. In this case it is greater than 5,which implies that we add 1 to the original digit or the rounding digit. Which means 0.4572, as 7 is greater than 5, we ass 1 to 5 which is our original digit. As a result we get 0.4672. last step, replace all the digit to the right of the one being rounded by zeros. So, 0.4682 we replace all the digits which is after 6 with zeros. So we get 0.4600 or 0.46.

The Process of Division of Decimal Numbers

The number systems are very interesting to study. Various arithmetic operations can be carried out on them. Multiplication is one of them. The reverse of this operation is division. Division is bit more different from addition and subtraction. The division of natural and whole numbers are almost similar. Any number divided by ‘0’ gives infinity. Infinity is a number which is very large and not defined. Zero on dividing by zero also gives infinity. This question was first raised by the great Indian mathematician Ramanujam.  Zero divided by any number gives zero. Any number divided by the number one gives the same number. This is the same with multiplication. Any number multiplied with one gives the same number. But any number multiplied by zero gives zero.

The process of dividing decimals with whole numbers is similar to the division of two whole numbers. The simple difference is in the placement of the decimal point. This is very important as the placing of decimal point can change the whole number. So, one has to be very careful about this. There can be dividing decimals problems in mathematics and have to be solved carefully. The decimals division is very much similar to the division of natural numbers or the whole numbers. The placement of the decimal point in the final answer is very crucial. Many dividing decimals examples can be used to explain the concept. The following steps to dividing decimals have to be followed to arrive at the final answer.

These are very easy steps.
If a decimal number is divided by a whole number, the division is carried out as usual. The digits present after the decimal point are noted and the decimal point is placed at the same point in the final answer. If the division is carried out between two decimal numbers then the decimal numbers are first converted into whole numbers by multiplying by 10, 100 or thousand and so on depending on how many digits are present after decimal point in the denominator. If the numbers to be divided are 8.68 and 1.2, then the numbers are multiplied by 10. The numbers now become 86.8 and 12. Now the division process is carried out. First ‘868’ is divided by 12. In the final answer decimal point is placed one point from the right. This is how the final answer is got.

Set Theory Rules are as follows

(A∪A) = A,  (A∩A) = A (idempotent law)
(A∪B)’ = A’∩B’, (A ∩ B) ′ = A ′ ∪ B ′ (De Morgan’s law)
A∪ B = B ∪ A, A∩B = B∩A (commutative law)
(A ∩ B) ∩ C = A ∩ (B ∩ C), (A ∪ B) ∪ C = A ∪ (B ∪ C) (associative law)
A ∪ φ = A, A ∩ U = A, A ∪ U = U, A ∩ φ = φ (identity law)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)   (distributive law)
(A’)’ = A (involution law)

Basic Set Theory Properties are:
A st is inherently unordered. This means that the order of elements in a st does not make a new st. Change in order of elements inside a st is feasible. For example if st A = {a, x, c} then A = {a, c, x} = {x, c, a} = {x, a, c} = {c, a, x} = {c, x, a}.
Each element in a st is distinct. Multiple repetition of an element makes no difference. For example if st A = {2, 3} then A = {2, 2, 3, 3} = {2, 3, 3} = {2, 2, 3} = {2, 2, 2, 3, 3, 3, 3} = {2, 3} and so on.
Above rules and properties of Set Theory Help you to solve Set Theory Questions easily.
Let us see some Set Theory Problems and Solutions now:

Q.1) prove that (A ′ ∩ B) ′ ∩ (A ∪ B) = A
Sol.) lets begin with LHS of the above problem:
LHS = ((A ′) ′ ∪ B ′) ∩ (A ∪ B) (by De Morgan’s law)
= (A∪ B’) ∩ (A ∪ B) (by involution law)
= A∪ (B∩ B’) (distributive law/property)
= A ∪ φ (complement law)
= A (identity law) = RHS
Hence, proved.

Q.2) in a community if 70 % people can speak English and 60% people can speak French then find the number of people who can speak both languages.
Sol.) Number of people who can speak English = n(E) = 70 %
Number of people who can speak French = n(F) = 60%
Number of people who can speak both languages will be n(E∩F) = ?
Total number of people in the community n (E∪F)= 100%
Now n (E∪F)= n(E) + n(F) - n(E∩F)
100=70 + 60 - n(E∩F)
100=130 - n(E∩F)
n(E∩F)=30%