• Title: "Discrete Mathematics and Its Applications" Third Edition
• Author: Kenneth H. Rosen

P181:
1. What rule of inference is used in each of the following arguments?
a) Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major.
Answer: Addition.
 
b) Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major.
Answer: Simplification.
 
c) If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed.
Answer: Modus ponens.
 
d) If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today.
Answer: Modus tollens.
 
e) If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn.
Answer: Hypothetical syllogism.
 

11. Prove that the sum of an irrational number and a rational number is irrational using a proof by contradiction.
Proof: Suppose that i is an irrational number, r is a rational number, and i+r is a rational number.
r is a rational number. => -r is also a rational number.
The sum of two rational numbers should be a rational number. => (i+r) + (-r) = i is a rational number. This is a contradiction with that i is an irrational number.
 

P198:
3. Use mathematical induction to prove that 3 + 3 * 5 + 3 * 5² + ... + 3 * 5ⁿ = 3(5ⁿ⁺¹ -1)/4 whenever n is a nonnegative integer.
Proof: Let P(n) be "3 + 3 * 5 + 3 * 5² + ... + 3 * 5ⁿ = 3(5ⁿ⁺¹ -1)/4", where n=0, 1, 2, ...

• Basis step: 3 = 3(5⁰ - 1)/4 => P(0) is true.
• Inductive step: Assume P(n) is true, i.e. 3 + 3 * 5 + 3 * 5² + ... + 3 * 5ⁿ = 3(5ⁿ⁺¹ -1)/4
Then 3 + 3 * 5 + 3 * 5² + ... + 3 * 5ⁿ + 3 * 5ⁿ⁺¹
= 3(5ⁿ⁺¹ -1)/4 + 3 * 5ⁿ⁺¹
= 3 * [ (5ⁿ⁺¹ -1)/4 + 5ⁿ⁺¹ ]
= 3 * ( 5 * 5ⁿ⁺¹ -1)/4
= 3 * (5ⁿ⁺² - 1)/4
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.

10. Prove that 1 * 1! + 2 * 2! + ... + n * n! = (n+1)! - 1 whenever n is a positive integer.
Proof: Let P(n) be " 1*1! + 2*2! + ... + n*n! = (n+1)! - 1 ", where n=1, 2, 3, ...

• Basis step: 1*1! = (1+1)! - 1 = 1 => P(1) is true.
• Inductive step: Assume P(n) is true, i.e. 1*1! + 2*2! + ... + n*n! = (n+1)! - 1
Then 1*1! + 2*2! + ... + n*n! + (n+1)*(n+1)!
= (n+1)! - 1 + (n+1)*(n+1)!
= (n+1)! * (1+ n+1) - 1
= (n+1)! * (n+2) - 1
= (n+2)! - 1
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.
 
 

• Basis step: 2⁵ = 32 > 5² = 25 => P(5) is true.
• Inductive step: Assume P(n) is true, i.e.  2ⁿ > n²
Then 2ⁿ⁺¹ = 2ⁿ * 2 > n² * 2
 (n+1)² = n² + 2n + 1
  2ⁿ⁺¹ - (n+1)²
= 2 * n² - (n² + 2n + 1)
= n² - 2n -1
= n -2n + 1 - 2
= (n-1)² -2
> 0 whenever n>4.
Such that 2ⁿ⁺¹ > (n+1)² whenever n>4.
The last inequality shows that P(n+1) is true. This completes the inductive step and completes the proof.
 

• Basis step: 1*2*3 = 1*(1+1)*(1+2)*(1+3)/4 = 6 => P(1) is true.
• Inductive step: Assume P(n) is true, i.e. 1*2*3 + 2*3*4 + ... + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4
Then 1*2*3 + 2*3*4 + ... + n(n+1)(n+2) + (n+1)*(n+2)(n+3)
= n(n+1)(n+2)(n+3)/4 + (n+1)(n+2)(n+3)
= (n+1)(n+2)(n+3) * (n/4 + 1)
= (n+1)(n+2)(n+3)(n+4)/4
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.

• Basis step: 1² = (-1)^1-1 * 1 * (1+1)/2 = 1 => P(1) is true.
• Inductive step: Assume P(n) is true, i.e. 1² - 2² + 3² - ... + (-1)^n-1 n² = (-1)^n-1n(n+1)/2
Then 1² - 2² + 3² - ... + (-1)^n-1 n² + (-1)ⁿ(n+1)²
= (-1)^n-1n(n+1)/2 + (-1)ⁿ(n+1)²
= (-1)ⁿ(n+1) * ( -n/2 + n+1 )
= (-1)ⁿ(n+1) * ( n/2+1 )
= (-1)ⁿ(n+1)(n+2)/2
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.

21. Use mathematical induction to show that 5 divides n⁵ - n whenever n is a nonnegative integer.
 Proof: Let P(n) be " 5 divides n⁵ - n ", where n = 0, 1, 2, ...

• Basis step: 5 devides  0⁵ - 0 = 0 => P(0) is true.
• Inductive step: Assume P(n) is true, i.e. 5 divides n⁵ - n.
Then (n+1)⁵ - (n+1)
= n⁵  + 5n⁴  + 10n³  + 10n²  + 5n + 1 - n - 1
= (n⁵ - n)+ 5n⁴ + 10n³  + 10n²  + 5n
= (n⁵ - n)+ 5*(n⁴ + 2n³  + 2n²  + n)
(n⁵ - n) can be divided by 5, apparently 5*(n⁴ + 2n³  + 2n²  + n) can be divided by 5. This means that P(n+1) is true. This completes the inductive step and completes the proof.

(A₁ U A₂ U ... U A_n) ^ B = (A₁ ^ B) U (A₂ ^ B) U ... U (A_n ^ B).

• Basis step: (A₁ U A₂)^ B = (A₁ ^ B) U (A₂ ^ B) by "Distributive laws" => P(2) is true.
• Inductive step: Assume P(n) is true, i.e. (A₁ U A₂ U ... U A_n) ^ B = (A₁ ^ B) U (A₂ ^ B) U ... U (A_n ^ B)
Then (A₁ U A₂ U ... U A_n U A_n+1) ^ B
= [ (A₁ U A₂ U ... U A_n) U A_n+1 ] ^ B
= [ (A₁ U A₂ U ... U A_n) ^ B ] U (A_n+1 ^ B)
= [ (A₁ ^ B) U (A₂ ^ B) U ... U (A_n ^ B) ] U (A_n+1 ^ B)
=  (A₁ ^ B) U (A₂ ^ B) U ... U (A_n ^ B) U (A_n+1 ^ B)
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.

40. Use mathematical induction to show that -(p₁ V p₂ V ... V p_n) is equivalent to -p₁ ^ -p₂ ^ ... ^ -p_n whenever p₁, p₂, ..., p_n are propositions. Where "V" means "OR", and "^" means "AND".
 Proof: Let P(n) be " -(p₁ V p₂ V ... V p_n) is equivalent to -p₁ ^ -p₂ ^ ... ^ -p_n "

• Basis step: -(p₁ V p₂) is equivalent to -p₁ ^ -p₂ by "De Morgan's laws" => P(2) is true.
• Inductive step: Assume P(n) is true, i.e. -(p₁ V p₂ V ... V p_n) is equivalent to -p₁ ^ -p₂ ^ ... ^ -p_n
Then -(p₁ V p₂ V ... V p_nV p_n+1) = -[ (p₁ V p₂ V ... V p_n) V p_n+1 ]
is equivalent to -(p₁ V p₂ V ... V p_n) ^ -p_n+1 
is equivalent to (-p₁ ^ -p₂ ^ ... ^ -p_n) ^ -p_n+1 
is equivalent to -p₁ ^ -p₂ ^ ... ^ -p_n ^ -p_n+1
The last "equivalent relation" shows that P(n+1) is true. This completes the inductive step and completes the proof.

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