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.