# The statement is true for ```1
MATHEMATICAL INDUCTION
Mathematical Induction is a powerful and elegant technique for proving certain types of mathematical
statements: general propositions which assert that something is true for all positive integers or for all
positive integers from some point on.
Why is this important?
Statement: If p is any prime number, 2p − 1 is also a prime. True? Not true?
p
2
3
5
7
2p − 1
3
7
31
127
Since 3, 7, 31, 127 are all primes, we may be satisﬁed the result is always
true. But if we try the next prime, 11, we ﬁnd that 211 − 1 = 2047 = 23 &times; 89.
So it is not a prime, and our general assertion is therefore FALSE.
General proof? Mathematical induction is one way of doing this.
Principle of Mathematical Induction (takes three steps)
TASK: Prove that the statement is true for all n∈ 𝑵
1. Check that the statement Pn is true for n = 1.
(Or, if the assertion is that the statement is true for n ≥ a, prove it for n = a.)
2. Assume that the statement is true for n = k (inductive hypothesis)
3. Prove that if the statement is true for n = k, then it must also be true for n = k + 1
Automatic “statement proving” machine: We checked the proposition for n = 1.
By the inductive step, since it is true for n = 1, it is also true for n = 2. Again, by the inductive step, since it is true
for n = 2, it is also true for n = 3. And since it is true for n = 3, it is also true for n = 4, and so on.
No recipe for how to do this. It varies from problem to problem, depending on the mathematical content. You have
to use your ingenuity, common sense and knowledge of mathematics here.
𝑛
EXAMPLE: Prove that for 𝑛 ≥ 1
∑ 𝑟𝑟! = (𝑛 + 1)! − 1
𝑟=1
1. 𝑛 = 1
1
∑ 𝑟𝑟! = 1 ∙ 1! = (1 + 1)! − 1 = 2 − 1 = 1
𝑟=1
𝑘+1
Statement is true for 𝑛 = 1
𝑔𝑜𝑎𝑙: 𝑝𝑟𝑜𝑣𝑒 ∑ 𝑟𝑟! = (𝑘 + 1 + 1)! − 1
2. Assume the statement is true for 𝑛 = 𝑘
𝑟=1
𝑘
𝑘
∑ 𝑟𝑟! = (𝑘 + 1)! − 1
𝑠𝑡𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝐿𝑆, 𝑎𝑛𝑑 𝑢𝑠𝑒 ∑ 𝑟𝑟! = (𝑘 + 1)! − 1
𝑟=1
𝑟=1
𝑘+1
3.
𝑖𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝐿𝑆 = 𝑅𝑆
𝑘
∑ 𝑟𝑟! = ∑ 𝑟𝑟! + (𝑘 + 1)(𝑘 + 1)! = (𝑘 + 1)! (1 + 𝑘 + 1) − 1 = (𝑘 + 2)! − 1
𝑟=1
𝑟=1
(𝑘 + 1)! − 1
The statement is true for 𝑛 = 1,
assuming that is true for 𝑛 = 𝑘, statement is true for 𝑛 = 𝑘 + 1
∴ 𝑆𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 1,
𝑏𝑦 𝑚𝑎𝑡ℎ 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝑄. 𝐸. 𝐷
2
1
EXAMPLE: P r o v e b y m a t h i n d u c t i o n t h a t the sum of the first positive integers is n(n + 1).
2
1. 𝑛 = 1
1
1 = 1(1 + 1) = 1
2
Statement is true for 𝑛 = 1
2. Assume the statement is true for 𝑛 = 𝑘
1
1 + 2 + ⋯ + 𝑘 = 𝑘(𝑘 + 1)
2
3.
1
1
1
1 + 2 + ⋯ + 𝑘 + (𝑘 + 1) = 𝑘(𝑘 + 1) + (𝑘 + 1) = (𝑘 + 1) ( 𝑘 + 1) = (𝑘 + 1)(𝑘 + 2)
2
2
2
1
𝑘(𝑘 + 1)
2
The statement is true for 𝑛 = 1,
assuming that is true for 𝑛 = 𝑘, statement is true for 𝑛 = 𝑘 + 1
∴ 𝑆𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 1,
𝑏𝑦 𝑚𝑎𝑡ℎ 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
EXAMPLE: For every positive integer n ≥ 2, n3 − n is a multiple of 6.
Statement: n3 – n = 6m 𝑚 ∈ 𝑁
1.
n = 2,
n3 − n = 23 − 2 = 6
which is a multiple of 6 (6 &times; 1).
Statement is true for 𝑛 = 2
2. Assume the statement is true for 𝑛 = 𝑘
k3 – k= 6m
for some integer m
3.
3
3
2
(k + 1) − (k + 1) = k + 3k + 3k + 1 − k − 1
3
2
= (k − k) + (3k + 3k)
= 6m + 3k(k + 1)
= 6m + 6ℓ
= 6(m + ℓ)
k(k + 1) is even number, k(k + 1)= 2ℓ
𝑚, ℓ ∈ 𝑁
The statement is true for 𝑛 = 2,
assuming that is true for 𝑛 = 𝑘, statement is true for 𝑛 = 𝑘 + 1
∴ 𝑆𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 2, 𝑏𝑦 𝑚𝑎𝑡ℎ 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝑄. 𝐸. 𝐷
𝑄. 𝐸. 𝐷
3
EXAMPLE: P r o v e t h a t 2𝑛 &gt; 𝑛2 for all natural numbers n ≥ 5.
1.
n = 5,
25 = 32 &gt; 𝑛2 = 25
Statement is true for 𝑛 = 5
2. Assume true for 𝑛 = 𝑘
2𝑘 &gt; 𝑘 2
3.
2𝑘+1 = 2(2𝑘 ) &gt; 2𝑘 2
o u r g o a l i s t o p r o v e 2𝑘+1 &gt; (𝑘 + 1)2
s o w e h a v e t o s h o w t h a t 2𝑘 2 &gt; (𝑘 + 1)2
o r 2𝑘 2 − 𝑘 2 + 2𝑘 + 1 &gt; 0
𝑘(𝑘 − 2) − 1 &gt; 0 𝑓𝑜𝑟 𝑘 ≥ 5
𝑄. 𝐸. 𝐷.
2𝑘+1 &gt; 2𝑘 2 &gt; (𝑘 + 1)2
𝑄. 𝐸. 𝐷
The statement is true for 𝑛 = 5,
assuming that it is true for 𝑛 = 𝑘, statement is true for 𝑛 = 𝑘 + 1
∴ 𝑆𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 5,
𝑏𝑦 𝑚𝑎𝑡ℎ 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
4
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