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Homework 5

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Module 9

    • Given that the definition of an odd number is 2𝑘+1, whereas an even number is 2𝑘, we can prove that the sum of two odd integers is even.
    • Given some odd integer, 2𝑛+1, and another 2𝑚+1, the sum is 2𝑛+1+2𝑚+1=2𝑛+2𝑚+2=2(𝑛+𝑚+1) which is of form 2𝑘. □
    • Let 𝑟 be some rational number and 𝑥 be some irrational number.
    • Suppose 𝑟+𝑥 is rational, this means that there exists some 𝑦 such that 𝑟+𝑥=𝑦, where 𝑦 is rational.
    • You can isolate 𝑥, 𝑥=𝑦−𝑟. Since 𝑦 and 𝑟 are both rational, the difference of two rational numbers is also rational. This would imply that 𝑥 is rational, but that contradicts the original assumption.
    • Since the assumption that 𝑟+𝑥 is a contradiction, 𝑟+𝑥 must be irrational.
    • ∴ The sum of an irrational number and a rational number is irrational. □
    • A rational number is defined as the ratio of two integers, such that 𝑟=𝑎𝑏, where 𝑎 and 𝑏 are integers, and 𝑏≠0.
    • Let 𝑛 and 𝑚 be two rational numbers, 𝑛=𝑎1𝑏1, and 𝑚=𝑎2𝑏2.
    • Multiplying 𝑛 and 𝑚 results in 𝑛×𝑚=𝑎1×𝑎2𝑏1×𝑏2.
    • Since the product of two integers is also an integer, 𝑛×𝑚 follows the form of 𝑎𝑏, where both 𝑎 and 𝑏 are integers.
    • ∴ The product of two rational numbers is always rational. □
    • A number is even if it follows the form 2𝑘, and odd if it follows the form 2𝑘+1.
    • Given that the product of 𝑚𝑛 is even, it is of the form 𝑚𝑛=2𝑘.
    • Assuming both are odd, 𝑚=2𝑎+1, and 𝑛=2𝑏+1.
    • 𝑚𝑛=(2𝑎+1)(2𝑏+1)=2𝑎×2𝑏+2𝑎+2𝑏+1=4𝑎𝑏+2𝑎+2𝑏+1 which is odd.
    • Since 𝑚𝑛 is odd if both 𝑚 and 𝑛 are odd, then at least one of 𝑚 or 𝑛 must be even for 𝑚𝑛 to be even. □
    • a.
      • The contrapositive is if 𝑛 is odd, then 𝑛3+5 is even.
      • Assuming 𝑛 is odd, 𝑛=2𝑘+1, 𝑛3=(2𝑘+1)3=8𝑘3+12𝑘2+6𝑘+1, which is odd.
      • 𝑛3+5 would therefore have to be even, since the sum of two odd numbers is even.
      • ∴ By contraposition, 𝑛 must be even give that 𝑛3+5 is odd.
    • b.
      • Suppose 𝑛3+5 is odd, and 𝑛 is also odd.
      • Suppose 𝑛 is odd, 𝑛=2𝑘+1,𝑛3=(2𝑘+1)3=8𝑘3+12𝑘2+6𝑘+1, which is odd.
      • 𝑛3+5 would be even, but the original statement was that 𝑛3+5 is odd, which is a contradiction.
      • ∴ We can conclude that 𝑛 must be even given that 𝑛3+5 is odd. □

Module 10

    • a.𝑃(1)=12=1(1+1)(2⋅1+1)6
    • b.1×2×36=66=1
    • c.12+22+…+𝑘2=𝑘(𝑘+1)(2𝑘+1)6,𝑘∈ℤ+
    • d.12+22+…+𝑘2+(𝑘+1)2=(𝑘+1)(𝑘+2)(2(𝑘+1)+1)6
    • e.12+22+…+𝑘2+(𝑘+1)2=(𝑘+1)(𝑘+2)(2(𝑘+1)+1)6=(𝑘+1)(𝑘+2)(2𝑘+3)6
      • Left Side: (12+22+…+𝑘2)+(𝑘+1)2
      • Substitute: 𝑘(𝑘+1)(2𝑘+1)6+(𝑘+1)2
      • Factor out (𝑘+1): 𝑘(𝑘+1)(2𝑘+1)6+6(𝑘+1)26=(𝑘+1)(𝑘(2𝑘+1)+6(𝑘+1))6
      • Expand: (𝑘+1)(2𝑘2+𝑘+6𝑘+6)6=(𝑘+1)(2𝑘2+7𝑘+6)6
      • Factor (2𝑘2+7𝑘+6): (𝑘+1)(𝑘+2)(2𝑘+3)6 which matches right-hand.
    • f. Since we have proven 𝑃(1) is true and that if 𝑃(𝑘) is true, then 𝑃(𝑘+1) is true, induction guarantees that the formula holds for all positive integers 𝑛. □
    • a.𝑃(1)=13=(1(1+1)2)2
    • b.1=(1×22)2=(22)2=12=1
    • c.13+23+…+𝑘3=(𝑘(𝑘+1)2)2,𝑘∈ℤ+
    • d.13+23+…+𝑘3+(𝑘+1)3=((𝑘+1)(𝑘+2)2)2
    • e.(13+23+…+𝑘3)+(𝑘+1)3
      • Substitute: (𝑘(𝑘+1)2)2+(𝑘+1)3
      • Factor out (𝑘+1)3: 𝑘2(𝑘+1)24+4(𝑘+1)34=(𝑘+1)2(𝑘2+4(𝑘+1))4
      • Expand: (𝑘+1)2(𝑘2+4(𝑘+1))4=(𝑘+1)2(𝑘2+4𝑘+4)4=(𝑘+1)2(𝑘+2)24
      • Simplify: ((𝑘+1)(𝑘+2)2)2 which matches right-hand.
    • f. Since we proved 𝑃(1) is true, and that if 𝑃(𝑘) is true, then 𝑃(𝑘+1) is true, induction guarantees that the formula holds for all positive integers 𝑛. □
    • Base Case (𝑛=0): (2(0)+1)2=12=1≡(0+1)(2(0)+1)(2(0)+3)3=1×1×33=33=1
    • Inductive Hypothesis: 12+32+52+…+(2𝑘+1)2=(𝑘+1)(2𝑘+1)(2𝑘+3)3,𝑘∈ℤ+
    • Inductive Step: 12+32+52+…+(2𝑘+1)2+(2(𝑘+1)+1)2=(𝑘+2)(2(𝑘+1)+1)(2(𝑘+1)+3)3
      • Substitute: (𝑘+1)(2𝑘+1)(2𝑘+3)3+(2𝑘+3)2
      • Expand: (𝑘+1)(2𝑘+1)(2𝑘+3)3+3(2𝑘+3)23=(𝑘+1)(2𝑘+1)(2𝑘+3)+3(2𝑘+3)23
      • Factor out (2𝑘+3): (2𝑘+3)((𝑘+1)(2𝑘+1)+3(2𝑘+3))3
      • Expand: (2𝑘+3)(2𝑘2+𝑘+2𝑘+1+6𝑘+9)3=(2𝑘+3)(2𝑘2+9𝑘+10)3
      • Factor: (2𝑘+3)(𝑘+2)(2𝑘+5)3
      • Since 2(𝑘+1)+1=2𝑘+3 and 2(𝑘+1)+3=2𝑘+5, the sides match
    • Since the formula holds for 𝑛=0, and that if the formula holds for 𝑛=𝑘, then it also holds for 𝑛=𝑘+1, by induction, the formula is true for all positive integers 𝑛. □
    • Base Case (𝑛=1): 1×1!=1≡(1+1)!−1=2!−1=2−1=1
    • Inductive Hypothesis: 1×1!+2×2!+…+𝑘×𝑘!=(𝑘+1)!−1,𝑘∈ℤ+
    • Inductive Step: 1×1!+2×2!+…+𝑘×𝑘!+(𝑘+1)×(𝑘+1)!=(𝑘+2)!−1
      • Substitute: (𝑘+1)!−1+(𝑘+1)×(𝑘+1)!
      • Factor out (𝑘+1)!: (𝑘+1)!−1+(𝑘+1)(𝑘+1)!=(𝑘+1)!(1+(𝑘+1))−1=(𝑘+1)!(𝑘+2)−1=(𝑘+2)!−1, which matches the formula
    • Since the formula holds for 𝑛=1, and that if the formula holds for 𝑛=𝑘, then it also holds for 𝑛=𝑘+1, by induction, the formula is true for all positive integers 𝑛. □
    • Base Case (𝑛=0): 3≡3(50+1−1)4=3(5−1)4=3×44=3
    • Inductive Hypothesis: 3+3×5+3×52+…+3×5𝑘=3(5𝑘+1−1)4,𝑘∈ℤ+
    • Inductive Step: 3+3×5+3×52+…+3×5𝑘=3(5𝑘+1−1)4
      • Substitute: 3(5𝑘+1−1)4+3×5𝑘+1
      • Factor out 3: 3(5𝑘+1−1)4+12×5𝑘+14=3(5𝑘+1−1)+12×5𝑘+14=3(5𝑘+1)−3+12(5𝑘+1)4=15(5𝑘+1)−34=3(5×5𝑘+1−1)4=3(5𝑘+2−1)4, which matches the formula
    • Since the formula holds for 𝑛=0, and that if the formula holds for 𝑛=𝑘, then it also holds for 𝑛=𝑘+1, by induction, the formula is true for all positive integers 𝑛. □
    • Base Case (𝑛=0): 2≡1−(−7)0+14=1+74=84=2
    • Inductive Hypothesis: 2−2×7+2×72−…+2(−7)𝑘=1−(−7)𝑘+14,𝑘∈ℤ+
    • Inductive Step: 2−2×7+2×72−…+2(−7)𝑘=1−(−7)𝑘+14
      • Substitute: 1−(−7)𝑘+14+2(−7)𝑘+1
      • Factor out 2(−7)𝑘+1: 1−(−7)𝑘+1+8(−7)𝑘+14=1−(−7)𝑘+1+8(−7)𝑘+14=1+7(−7)𝑘+14=1−(−7)𝑘+24, which matches the formula
    • Since the formula holds for 𝑛=0, and that if the formula holds for 𝑛=𝑘, then it also holds for 𝑛=𝑘+1, by induction, the formula is true for all positive integers 𝑛. □

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