## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 1 - Problem Solving and Critical Thinking - 1.1 Inductive and Deductive Reasoning - Exercise Set 1.1 - Page 13: 71

#### Answer

a. The sums are all equal to 30. b. The sums are all equal to 36. c. If we choose $a=4$, $b=1$, $c=2$, the sums are all equal to 12. d. We may inductively conclude that all the sums are all equal to $3*a$. e. When we sum up each row, column and diagonal without replacing the variables, we can deductively conclude that all the sums are equal to $3*a$.

#### Work Step by Step

* The table is provided in the picture * a) $a + b = 10 + 6 = 16$ $a -b-c=10-6-1=3$ $a+c=10+1=11$ Row #1: 16 , 3 , 11 $a-b+c=10-6+1=5$ $a=10$ $a+b-c=10+6-1=15$ Row #2: 5 , 10 , 15 $a-c = 10-1=9$ $a+b+c=10+6+1=17$ $a-b=10-6=4$ Row #3: 9 , 17 , 4 The result table is: 16, 3, 11 5, 10, 15 9, 17, 4 The sums are: Row #1: $16 + 3 + 11 = 30$ Row #2: $5 + 10 + 15 = 30$ Row #3: $9 + 17 + 4 = 30$ Column #1: $16 + 5 + 9 = 30$ Column #3: $3 + 10 + 17 = 30$ Column #3: $11 + 15 + 4 = 30$ Diagonal #1: $16 + 10 + 4 = 30$ Diagonal #2: $9 + 10 + 11 = 30$ We observe that all the sums are equal to 30. b. repeat part (a) for $a=12$, $b=5$, and $c=2$. $a + b = 12 + 5 = 17$ $a -b-c=12-5-2=5$ $a+c=12+2=14$ Row #1: 17 , 5 , 14 $a-b+c=12-5+2=9$ $a=12$ $a+b-c=12+5-2=15$ Row #2: 9 , 12 , 15 $a-c = 12-2=10$ $a+b+c=12+5+2=19$ $a-b=12-5=7$ Row #3: 10 , 19 , 7 The result table is: 17, 5, 14 9, 12, 15 10, 19, 7 The sums are: Row #1: $17 + 5 + 14 = 36$ Row #2: $9 + 12 + 15 = 36$ Row #3: $10 + 19 + 7 = 36$ Column #1: $17 + 9 + 10 = 36$ Column #2: $5 + 12 + 19 = 36$ Column #3: $14 + 15 + 7 = 36$ Diagonal #1: $17 + 12 + 7 = 36$ Diagonal #2: $10+ 12 + 14 = 36$ We observe that all the sums are equal to 36. c. repeat part (a) for values $a$, $b$, and $c$ of your choice. $a = 4$, $b = 1$, $c = 2$ $a + b = 4+1 = 5$ $a -b-c=4-1-2=1$ $a+c=4+2=6$ Row #1: 5 , 1 , 6 $a-b+c=4-1+2=5$ $a=4$ $a+b-c=4+1-2=3$ Row #2: 5 , 4 , 3 $a-c = 4-2=2$ $a+b+c=4+1+2=7$ $a-b=4-1=3$ Row #3: 2 , 7 , 3 The result table is: 5, 1, 6 5, 4, 3 2, 7, 3 The sums are: Row #1: $5 +1+6 = 12$ Row #2: $5+4+3 = 12$ Row #3: $2+7+3 = 12$ Column #1: $5+5+2 = 12$ Column #2: $1+4+7 = 12$ Column #3: $6+3+3 = 12$ Diagonal #1: $5+4+3= 12$ Diagonal #2: $2+4+6= 12$ We observe that all the sums are equal to 12. d. We may inductively conclude that they are all equal to $a*3$ e. If we sum up each row, column and diagonal without replacing the variables, we can get the results in variable form, which could deductively prove our conjecture. Row #1: $a+b+a-b-c+a+c=a+a+a=3a$ Row #2: $a-b+c+a+a+b-c=a+a+a=3a$ Row #3: $a-c+a+b+c+a-b=a+a+a=3a$ Column #1: $a+b+a-b+c+a-c=a+a+a=3a$ Column #2: $a-b-c+a+a+b+c=a+a+a=3a$ Column #2: $a+c+a+b-c+a-b=a+a+a=3a$ Diagonal #1: $a+b+a+a-b=a+a+a=3a$ Diagonal #2: $a-c+a+a+c=a+a+a=3a$ We used deductive reasoning to prove that all the sums are equal to $3a$.

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