- What is Deductive reasoning
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- What is Inductive reasoning
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- The Principle of Mathematical Induction<
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- Solved Examples

Deductive reasoning" refers to the process of concluding that something must be true because it is a special case of a general principle that is known to be true. For example, if you know the general principle that the sum of the angles in any triangle is always 180 degrees, and you have a particular triangle in mind, you can then conclude that the sum of the angles in your triangle is 180 degrees.

Deductive reasoning is logically valid and it is the fundamental method in which mathematical facts are shown to be true.

Please dont confuse inductive reasoning with "mathematical induction" or and "inductive proof", which is something quite different

(

(

+ 1, i.e.,

(

Here the Step 1 is Basis of Induction

Step 2 is called inductive hypotthesis. Here we are assuming the identity to be true as Step 1 is true(special case)

Step 3 is called inductive step. Here we prove the identity is true for k+1 on the basis of inductive hypothesis

So it is a deductive reasoning technique.

Let us assume this statement as P(n) Then P(0), we have LHS side as 0 RHS= 0(0+1)/2=0 So LHS=RHS

Let us assum P(n) is true,then

Now we have to prove for P(n+1)

0+1+2+......(n)+(n+1)=\frac{(n+1)[(n+1)+1]}{2}

Now lets us take LHS

= $\frac{n(n+1)}{2}+(n+1)$ as P(n) is assumed to be true

=$\frac{(n+1)[(n+1)+1]}{2}$

So P(n+1) is true if P(n) is true. Now Since P(0) is true, we can say P(n) is good for all natural numbers

To prove by mathematical induction that

6

is divisible by 43 for each positive integer n.

Let P(n) be the statement

P(n)=6

Basis of Induction

P(1) =6

= 559 = 43 X 13,

the formula is true for n = 1.

Inductive Hypothesis

Assume that P(n) is true for n = k, that is

P(k)=6

for some integer x.

Inductive Step

Now show that the formula is true for n = k + 1.

Observe that

P(k+1)=6

=

=6

=6X6

=6X6

= 6 X ( P(k) ) + 43 X 7

Since each component of this sum is divisible by 43 so is the entire sum and the formula holds for k + 1

Prove using Mathematical induction for $n\geqslant 1$

Let us assume this statement as P(n)

Then P(1)

$1=\frac{1(3-1)}{2}$

Inductive Hypothesis

Assume that P(n) is true for n = k, that is

$1+4+7+......(3k-2)=\frac{k(3k-1)}{2}$

Inductive Step

Now show that the formula is true for n = k + 1.

Observe that

$1+4+7+.....[3(k+1)-2]=\frac{(k+1)[3(k+1)-1])}{2}$

Let us take the LHS

$1+4+7+.....[3(k+1)-2]=1+4+7+......+(3k-2)+(3k+1)$

$=\frac{k(3-1)}{2}+(3k+1)$

$=\frac{3k^{2}+5k+2}{2}$

$=\frac{(k+1)[3(k+1)+1]}{2}$

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