Inductive Reasoning

Show your work for each problem.

1.
Which of these is an example of inductive reasoning?
1. a)If x + 3 = 7, then x = 4.
2. b)All squares have 4 sides. This shape is a square, so it has 4 sides.
3. c)Every multiple of 10 ends in 0. 70 is a multiple of 10, so 70 ends in 0.
4. d)The bus was late every morning this week, so it will probably be late tomorrow.
Answer: The bus was late every morning this week, so it will probably be late tomorrow.
The bus example uses several specific observations (this week's mornings) to predict what will probably happen. The other three start from a general rule or an equation and reach a conclusion that must be true: that is deductive reasoning.
2.
Use the conjecture "the sum of the first $n$ odd numbers is $n^2$ " to find $1 + 3 + 5 + 7 + 9 + 11$ without adding term by term.
Answer: 36
There are $6$ odd numbers, so the conjecture gives $6^2 = 36$ . Adding them confirms it: $1 + 3 + 5 + 7 + 9 + 11 = 36$ .
3.
Dots are arranged in triangles: figure 1 has $1$ dot, figure 2 has $3$ dots, figure 3 has $6$ dots and figure 4 has $10$ dots. If the pattern continues, how many dots does figure 6 have?
Answer: 21
1. How many dots are added from figure 3 to figure 4? 4
2. The added dots go $2, 3, 4$ . How many dots does figure 5 have? 15
3. How many dots does figure 6 have? 21
4.
Which number is a counterexample to the conjecture "all prime numbers are odd"?
1. a)2
2. b)9
3. c)15
4. d)7
Answer: 2
$2$ is prime (its only divisors are $1$ and $2$ ) and it is even, so it breaks the conjecture. $9$ and $15$ are odd but not prime, and $7$ is prime and odd, so none of them is a counterexample.
5.
Conjecture: "doubling a number always makes it bigger." Which number is equal to its own double and is therefore a counterexample?
Answer: 0
$2 \times 0 = 0$ , so doubling $0$ does not make it bigger. One counterexample is enough to show the conjecture is false. (Negative numbers also break it: $2 \times (-3) = -6$ , which is smaller.)
6.
A sunflower is $2$ cm tall after week 1, $5$ cm after week 2, $8$ cm after week 3 and $11$ cm after week 4. If the pattern continues, how tall will it be after week 8 (in cm)?
Answer: 23
1. How many centimeters does the sunflower grow each week? 3
2. How tall is it after week 5? 14
3. How tall is it after week 8? 23
7.
Conjecture: the sum of the first $n$ odd numbers is $n^2$ . Test it with $n = 5$ , then use it to predict the sum of the first $20$ odd numbers.
Answer: 400
1. What is $1 + 3 + 5 + 7 + 9$ ? 25
2. $25$ is the square of which number? 5
3. Using the conjecture, what is the sum of the first 20 odd numbers? 400
8.
Tom adds $1 + 5$ , $9 + 3$ and $15 + 7$ , gets an even number each time and concludes: "the sum of two odd numbers is always even." Which kind of reasoning did Tom use?
1. a)Circular reasoning
2. b)Mathematical induction
3. c)Inductive reasoning
4. d)Deductive reasoning
Answer: Inductive reasoning
Tom went from three specific cases to a general conclusion: that is inductive reasoning. His conjecture happens to be true, but three examples do not prove it. Mathematical induction, despite its name, is a deductive proof method.
9.
A triangle has $0$ diagonals, a quadrilateral $2$ , a pentagon $5$ , a hexagon $9$ , a heptagon $14$ and an octagon $20$ . Continue the pattern: how many diagonals does a nonagon (9 sides) have?
Answer: 27
The differences grow by $1$ : the next one is $7$ , so $20 + 7 = 27$ . The formula $\frac{n(n - 3)}{2}$ agrees: $\frac{9 \times 6}{2} = 27$ .
10.
Look at these products: $11 \times 11 = 121$ , $111 \times 111 = 12321$ , $1111 \times 1111 = 1234321$ . Make a conjecture and predict $11111 \times 11111$ .
Answer: 123454321
1. What is the middle (largest) digit of $1234321$ ? 4
2. With five ones, what should the middle digit be? 5
3. Write the predicted result of $11111 \times 11111$ . 123454321
11.
Lena claims: "squaring a number always makes it bigger." Test her claim with $\frac{1}{2}$ : what is $\left(\frac{1}{2}\right)^2$ ? Write your answer as a fraction.
Answer: 1/4
$\left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , and $\frac{1}{4} < \frac{1}{2}$ . So $\frac{1}{2}$ is a counterexample: squaring a number between $0$ and $1$ makes it smaller.
12.
The expression $n^2 + n + 41$ gives a prime number for every $n$ from $0$ to $39$ . For $n = 40$ it gives $1681$ . Which statement is correct?
1. a)The conjecture is proven because it worked for n = 0 to 39
2. b)One counterexample is not enough to reject the conjecture
3. c)1681 is prime, so the conjecture still holds
4. d)1681 = 41 × 41, so n = 40 is a counterexample
Answer: 1681 = 41 × 41, so n = 40 is a counterexample
$40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41 \times 41$ , which is not prime. So $n = 40$ is a counterexample, and one counterexample is enough to show that "always prime" is false, no matter how many cases worked before.
13.
At a party everyone shakes hands with everyone else exactly once. With $2$ people there is $1$ handshake, with $3$ people $3$ handshakes and with $4$ people $6$ handshakes. How many handshakes are there with $10$ people?
Answer: 45
1. The differences are $2$ and $3$ . How many handshakes are there with $5$ people? 10
2. How many handshakes are there with $6$ people? 15
3. The conjecture $\frac{n(n - 1)}{2}$ fits all these cases. Use it for $n = 10$ . 45
14.
Test the conjecture " $n^2 + n + 41$ is always prime" step by step and find the value that breaks it at $n = 40$ .
Answer: 1681
1. What is $n^2 + n + 41$ for $n = 1$ ? 43
2. What is $n^2 + n + 41$ for $n = 5$ ? 71
3. What is $n^2 + n + 41$ for $n = 40$ ? 1681
4. What is $1681 \div 41$ ? 41
15.
From the cases $0, 2, 5, 9$ (polygons with $3$ to $6$ sides) we made the conjecture that a polygon with $n$ sides has $\frac{n(n - 3)}{2}$ diagonals. Use it: how many diagonals does a polygon with $12$ sides have?
Answer: 54
$\frac{12 \times (12 - 3)}{2} = \frac{12 \times 9}{2} = \frac{108}{2} = 54$ . This conjecture has also been proven deductively: from each of the $n$ corners you can draw $n - 3$ diagonals, and each diagonal is counted twice.
16.
Sam notices: $2^2 - 1 = 3$ , $2^3 - 1 = 7$ , $2^5 - 1 = 31$ , $2^7 - 1 = 127$ , all prime. He conjectures: "if $p$ is prime, then $2^p - 1$ is prime." Which value of $p$ is a counterexample?
1. a)p = 11
2. b)p = 17
3. c)p = 13
4. d)p = 19
Answer: p = 11
$2^{11} - 1 = 2047 = 23 \times 89$ , which is not prime, so $p = 11$ breaks the conjecture. The others give primes: $2^{13} - 1 = 8191$ , $2^{17} - 1 = 131071$ , $2^{19} - 1 = 524287$ . Four supporting cases were not enough.
17.
Put points on a circle and join every pair of points with a straight line. Count the regions inside the circle: $1$ point gives $1$ region, $2$ points give $2$ , $3$ points give $4$ , $4$ points give $8$ , $5$ points give $16$ . A careful drawing with $6$ points, placed so that no three lines meet at one point, shows $31$ regions. Test the doubling conjecture: by how many regions does its prediction miss?
Answer: 1
1. Each count is multiplied by which number to get the next one? 2
2. What does the doubling conjecture predict for $6$ points? 32
3. The drawing shows $31$ regions. By how many regions does the prediction miss? 1
18.
Conjecture: "the sum of three consecutive whole numbers is always $3$ times the middle number." Test it inductively, then use the deductive version to find $49 + 50 + 51$ .
Answer: 150
1. What is $4 + 5 + 6$ ? 15
2. What is $15 \div 3$ ? (Compare it with the middle number.) 5
3. Test another case: what is $20 + 21 + 22$ ? 63
4. Deductively: $(n - 1) + n + (n + 1) = 3n$ . Use it to find $49 + 50 + 51$ . 150