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What is Deductive reasoning

Deductive reasoning also deductive logic or logical deduction or, informally, "top-down" logic is the process of reasoning from one or more statements (premises) to reach a logically certain conclusion
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.

What is Inductive reasoning

It is the process of reasoning that a general principle is true because the special cases you've seen are true. For example, if all the people you've ever met from a particular town have been very strange, you might then say "all the residents of this town are strange". That is inductive reasoning: constructing a general principle from special cases. It goes in the opposite direction from deductive reasoning.

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

The Principle of Mathematical Induction

Suppose there is a given statement P(n) involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true, and
(ii) If the statement is true for n = k (where k is some positive integer), then
iii) the statement is also true for n = k
+ 1, i.e., truth of P(k) implies the
truth of P
(k + 1).
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.

Solved Examples:

Given n belongs to the set of Natural Numbers. Prove that using principle of Mathematical induction
Solutions:
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

Example

To prove by mathematical induction that

6 (n+2) + 7(2n+1)
is divisible by 43 for each positive integer n.

Solution
Let P(n) be the statement
P(n)=6 (n+2) + 7(2n+1)
Basis of Induction
P(1) =6 (1+1) + 7(2+1)
= 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 (k+2) + 7(2k+1)=43x
for some integer x.

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

Observe that
P(k+1)=6 (k+1+2) + 7(2k+2+1)

= (k+3) + 7(2k+3)
=6(k+2) + 727(2k+1)
=6X6(k+2) + (43+6)X7(2k+1)
=6X6(k+2) + 6X7(2k+1)+43X7(2k+1)
= 6 X ( P(k) ) + 43 X 7(2k+1)

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

Prove using Mathematical induction for $n\geqslant 1$ $1+4+7+......(3n-2)=\frac{n(3n-1)}{2}$
Solution
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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