A property known to hold for one number holds for all natural numbers. Let N = {1,2,3,4,…} be the set of natural numbers, and P(n) be a mathematical statement involving the natural number n belonging to N such that
(i) P(1) is true, i.e., P(n) is true for n = 1.
(ii) P(n+1) is true whenever P(n) is true, i.e., P(n) is true implies that P(n+1) is true.
Then P(n) is true for all natural numbers n.
For example, we can prove by induction that all positive integers of the form 2n − 1 are odd. Let P(n) represent ‘2n − 1 is odd’.
(i) For n = 1, 2n − 1 = 2(1) − 1 = 1, and 1 is odd, since it leaves a remainder of 1 when divided by 2. Thus P(1) is true.
(ii) For any n, if 2n − 1 is odd (P(n)), then (2n − 1) + 2 must also be odd, because adding 2 to an odd number results in an odd number. But (2n − 1) + 2 = 2n + 1 = 2(n+1) − 1, so 2(n+1) − 1 is odd (P(n+1)). So P(n) implies P(n+1). Thus 2n − 1 is odd, for all positive integers n.
“You want a proof? OK, then.
A property known to hold for one number holds for all natural numbers. Let N = {1,2,3,4,…} be the set of natural numbers, and P(n) be a mathematical statement involving the natural number n belonging to N such that
(i) P(1) is true, i.e., P(n) is true for n = 1.
(ii) P(n+1) is true whenever P(n) is true, i.e., P(n) is true implies that P(n+1) is true.
Then P(n) is true for all natural numbers n.
For example, we can prove by induction that all positive integers of the form 2n − 1 are odd. Let P(n) represent ‘2n − 1 is odd’.
(i) For n = 1, 2n − 1 = 2(1) − 1 = 1, and 1 is odd, since it leaves a remainder of 1 when divided by 2. Thus P(1) is true.
(ii) For any n, if 2n − 1 is odd (P(n)), then (2n − 1) + 2 must also be odd, because adding 2 to an odd number results in an odd number. But (2n − 1) + 2 = 2n + 1 = 2(n+1) − 1, so 2(n+1) − 1 is odd (P(n+1)). So P(n) implies P(n+1). Thus 2n − 1 is odd, for all positive integers n.
How’s that for a proof?!" (◔_◔)