|
In Mathematical induction we find whether a given statement is true for all the natural Numbers or not. So the answer to What is Mathematical Induction is that it is basically a technique used to prove a statement or a theorem, or a formula that is advanced about all natural numbers. As we know that in the Natural Numbers only positive numbers are included. This should not be construed as a form of inductive reasoning. The simplest form of mathematical induction identify that whether a statement which include a natural number n holds for all values of n is true or not. The Proof of Mathematical Induction contains two steps, and the steps are shown below: |
The simplest form of mathematical induction identify that whether a statement which is included is a natural number n which holds for all values of n is either true or not. The Proof of Mathematical Induction contains two steps which are shown below:
1. The basic case of induction mathematical is used to show whether a statement holds the value of n is equal to the lowest value of n. Generally, we take the value of n as 0 or 1.
2. The inductive case of mathematical induction is used to show whether the given statement holds the value of n then the full statement also holds when the value of n is substituted to n + 1.
This given method is applicable by proving the statement is true for the starting value. Then we have to prove the process which is used to go from one value to the other value which is valid. If both the given values are taken, then any value can be calculated by performing the process.
Let us use the principle of mathematical induction is to prove the following statement:
When we have A(n) = 0 + 1 + 2 + 3 + …… + n = n (n + 1) / 2;
Assume that we have the value of n is ‘1’, then putting the value of n in the formula we get:
= n (n + 1) / 2;
n = 1;
A (1) = 1 (1 + 1) / 2;
So we get the value of left and right hand side same.
In Mathematical Induction we will find whether a given statement is true for natural Numbers or not. We know that Natural Numbers consist of positive numbers only. We can not plot this as it in form of inductive reasoning. Simplest form of mathematical induction is to classify whether a statement which includes a natural number ‘n’ holds for all values of 'n' to find whether they are true or not. Proof of mathematical induction has two steps, and steps are mention below:
1. First is basic case of induction mathematics that is used to show that statement holds value of 'n' which is equals to lowest value of 'n', basically we take value of 'n' as 0 or 1.
Second is inductive case of mathematical induction that is used to show that given statement holds value of 'n' and statement also holds true when we put value of 'n' as n + 1.
Now we will discuss mathematical induction inequality. As we know that if ‘<, >, ≥, ≤' these symbols are present in expression then we can say expression contains inequality. Mathematical induction inequality can be understood with help of an example. Suppose we have an inequality equation S (f) , f < 2f for any arbitrary value f > 0.
Here we will try to prove that q (f + 1) is true when q (f) is true.
= q (f + 1) is (f + 1) < 2(f + 1),
If we add 1 to both sides of q (f) then we obtain:
= f + 1 < 2(f) + 1.
So left side of inequality is correct because it matches left side of q (f + 1), now we need to solve right side.
We are trying to show that:
2(f) + 1 < 2(f + 1).
Therefore:
= f + 1 < 2(f) + 1 ≤ 2(f) + 2(f) = 2 * 2(f) = 2(f + 1).
Strong Mathematical Induction is an algebraic method to prove some statement or expression true by using contradiction method. According to this method we assume a hypothesis in starting which has to be proved wrong by the end of complete calculation. So to prove your induction to be true considered statement must be satisfied by at least one number. Each step of Mathematical Induction Problems are to be proved using mathematical formulas.
Let us consider an example of mathematical induction as i² >= 2i where i = 2, 3, 4 and so on. To start with consider your induction hypothesis that you want to prove. If i² >= 2i is true then it should be true for i = k also, where k = 2, 3, 4 and so on. So, we can write above equation as:
k² >= 2k. Therefore if it is true for i = k we must now prove our hypothesis true for i = k + 1 also. This is the step where you actually prove your hypothesis wrong by contradiction by performing certain mathematical calculations. So, problem can be written as:
i = k + 1 then (k + 1)² >= 2 (k + 1) for every (k + 1) = 2, 3, 4 and so on.
(k + 1)² >= 2 (k + 1),
k² + 2k + 2 >= 2k + 2,
We know k = i and i² = 2i so k² = 2k.
2k + 2k + 2 >= 2k + 2,
We know 2k > 1 because k > 1,
2k + 2k + 1 > 2k + 2.
Thus we see here that left side comes out to be greater than right side which proves our induction by contradicting the situation.
Mathematical Induction is an algebraic method to show that a statement or a postulate is true by using the contradiction technique. According to this method we assume a hypothesis in the starting which has to be proved wrong by the end of complete calculation. So to prove your induction to be true, the considered statement must be satisfied by at least one number. Each step of Mathematical Induction problems are to be proved using mathematical formulas.
Let us consider an example of mathematical induction as H² >= 2H where H = 2, 3, 4 and so on. Mathematical induction steps for this problem: To start with consider your induction hypothesis that you want to prove. If H² >= 2H is true then it should be true for H = k also, where k = 2, 3, 4 and so on. So, we can write above equation as:
k² >= 2k. Therefore if it is true for H = k we must now prove our hypothesis true for H = k + 1 also. This is the step where you actually prove your hypothesis wrong by contradiction by performing certain mathematical calculations. So, problem can be written as:
i = k + 1 then (k + 1)² >= 2 (k + 1) for every (k + 1) = 2, 3, 4 and so on.
(k + 1)² >= 2 (k + 1),
k² + 2k + 2 >= 2k + 2,
We know k = i and i² = 2i so k² = 2k.
2k + 2k + 2 >= 2k + 2,
We know 2k > 1 because k > 1,
2k + 2k + 1 > 2k + 2,
Thus we see here that left side comes out to be greater than right side which proves our induction by contradicting situation.