# Substitution inquiry

# The prompt

Mathematical inquiry processes: Generate examples and counter-examples; conjecture and reason; change conditions. Conceptual field of inquiry: Substitution of different types of numbers into expressions.

The prompt was devised by Mark Greenaway (an Advanced Skills Teacher in Suffolk, UK). In his original version, the prompt included values for a and b (2 and 3 respectively). In the prompt’s more open form (as shown above), students might suggest their own values to substitute into the inequality in order to develop conjectures about the relationship between a and b.

To date, students’ questions and observations about the prompt have included:

What are they trying to find out?

What values of a and b satisfy the inequality?

What does the sign mean?

a could be 2 and b could be 4.

This can only work when a ≥ 2 and b ≥ 3 with b > a.

What if a and b are negative numbers? Would the inequalities be reversed?

What if a and b are fractions?

Is it possible to change the order of the terms and expressions and find the same values of a and b that satisfy all permutations?

What if we changed the operations?

Could we use square roots?

This one must work all the time: a/b < a - b < a + b < ab.

Is it possible for a2, a + b, ab, b2 to be equal?

Inquiries have often led students into changing the order of the terms in the inequality. Other pathways include substituting decimal and negative numbers into the inequality, and using more complex expressions.

# Conjectures and result

## Conjectures

Five examples of the many conjectures about the relationship between a and b that have arisen in classroom inquiry are:

If a and b are consecutive whole numbers, then the inequality will never work. By making a = n and b = n + 1, a year 9 student showed how the four parts of the inequality become n2 < 2n + 1 < n2 + n < n2 + 2n + 1. The conjecture turns out to be correct, except when n = 2.

The values of a and b must be in the same times table to make this correct. A counter-example (like those in the table below) show this conjecture to be false.

As long as a is smaller than b, the prompt will work. This can be shown to be false with a counter-example, such as a = 3, b = 5.

If a = 1, then the inequality will never work. This is true because when a = 1, ab < a + b.

The inequality will not work if both numbers are negative. This is also true because if a and b are negative, a2 > a + b.

## Result of the inquiry

One notable result came from a group of students who decided to find the lowest value of b when a is set as a particular whole number.

The students explained their results by focusing on a2 and a + b. In order for a + b > a2, b = a2 - a + 1.

# Excitement and commitment

These are the questions and observations of Shawki Dayekh's year 9 class at Haverstock School in Camden (London, UK). The class had low prior attainment in mathematics. Shawki reflects on how the inquiry developed over two one-hour lessons:

"I didn’t expect the level of some of the questions from the class. Students were so excited to prove or disprove each other’s comments, observations and conjectures. One student, who was disengaged at the beginning of the inquiry, asked 'How long will it take to work out?' Well, I said it can take a lifetime if you want! Then the students dived into the problem. In the second lesson they went straight to making generalisations from their tables of values. One group wrote, 'If a and b are both positive and b > a, then the inequality will always work.' Another group, referring to their table of results, said, 'The inequality fails if a and b are both negative.' I have to say that I didn’t expect that level of commitment and mathematical language from the class. It was one of the best lessons I have had with them!"

Shawki was so enthusiastic about the depth of learning during the inquiry that he decided to extend it into the third lesson. Describing the process as "zooming in on a mathematical property", Shawki gave the class the following conjectures to test:

(1) If a < b, then a2 < b2

(2) If a < b, then a - b < 0

(3) If a and b are integers, then a + b < ab

(4) If 0 < a < 1 and b > 1, then ab > b

Shawki reports that all the students tested and wrote about one conjecture and many tackled two or three. Sofian, one of the students, even suggested revising the first conjecture to see if it was true for a3 and b3. Tasnim, another student, tested the first three conjectures and her results are pictured below. She shows that conjecture (1) is not always true when a is negative; conjecture (2) is always true; and conjecture (3) is true when both a and b are negative or positive, but not if one is negative and the other positive.

# Question-driven inquiry

These are the questions of a year 9 class (a middle set) in a UK comprehensive school. The students are curious to see what happens when they change the prompt. In order to help students decide on their own line of inquiry, the teacher produced this list of questions for the second lesson. In the pictures below, you can see students' inquiries into questions 2, 4 and 6.

# Adapting the prompt

This prompt was devised for a year 9 class with high prior attainment. The class had carried out mathematical inquiries before and were beginning to inquire independently. Students came up with a wide range of questions and observations that the teacher or, in the case of the example with a = 2 and b = 5, a student wrote on the board (see below). Another student explained that as b2/a is greater than a2/b, b must be greater than a.

The class was then given the opportunity to select a regulatory card to decide how the inquiry should proceed. Most students chose to Try to find more examples by which they meant to substitute more values for a and b into the inequality.

A minority, however, decided to determine the truth of the inequality. One table of four students wanted to Use a worksheet - a card they had selected on previous occasions - even though it was not included in the set. The teacher gave those students the structured inquiry sheet (see in the Resources section below).

As they explored, students began to appreciate that it was difficult to find values for a and b that satisfy the central inequality. One pair of girls proved that the middle two expressions should be reversed by creating two equivalent fractions with the denominator ab and using the inequality b > a to show that ab + b2 > a2 + b2. Another student proved the prompt was false by showing the middle inequality contradicted the condition b > a. He presented his proof (above) to the class at the end of the lesson. If the middle inequality is correct, he argued, then b < a, which we know to be false.

# Introducing substitution through inquiry

Rachel Mahoney, a mathematics teacher at Carre's Grammar School in Sleaford (Lincolnshire, UK), posted this picture on twitter. It shows the questions and observations from Rachel's year 7 class. The students who were carrying out only their second inquiry have already begun to suggest changes to the prompt, such as changing the inequality to greater than, changing the indices and extending the 'sequence'. Rachel reports that the prompt is "a great way to introduce substitution."

# Students' rich questions

Amanda Kirby a teacher of mathematics and the Numeracy Across the Curriculum Coordinator at St Clement Danes School, Hertfordshire (UK), used the substitution prompt for an observation lesson with her year 10 foundation GCSE class. She reports that the students were "fabulous at asking questions and questioning each other" (see their initial comments and questions below). The class quickly realised that the prompt is never true if a and b are negative numbers because, in that case, a2 > a + b. As the inquiry progressed, students created their own inequality: a - b2 < a2 - b. They wondered if a has to be greater than b for the inequality to be true. Overall, Amanda was pleased with the way the classroom was 'buzzing' during the inquiry and with students' willingness to challenge each other's reasoning.

# Lines of inquiry

The pictures below show the inquiries of year 9 students. In their questions, the students are wondering about the values of a and b that satisfy the inequality and considering if there is one pair of values or more. They are also thinking about how to find the values by considering different types of numbers. There is speculation about the relationship between a and b and, finally, a suggestion to swap the order of the terms and expressions that has the potential to open a new inquiry pathway. The next two pictures show examples of inquiry pathways. One student has reached conclusions about the possibility of using different types of numbers, while another has calculated the lowest value of b for each value of a and then devised the formula: Lowest value of b for a given a = n(n - 1) + 1 (where n = a).

The pictures were posted on twitter by the students' teacher Aine Carroll. Aine reports that, "Inquiry Maths is becoming a core part of my year 9 lessons."

# Extending the inquiry

The extension to the inquiry developed out of a student's question about the original prompt. She asked, "Are there values of a and b to make all the different orders of the inequality?" As there are 24 permutations of the four parts of the inequality, it might be better to start with three terms or expressions. In that case, there are six permutations. In the example (illustration above), we have:

b - a < b/a < ab b - a < ab < b/a

b/a < b - a < ab ab < b - a < b/a

b/a < ab < b - a ab < b/a < b - a

In this PowerPoint designed for online inquiry there are four challenges. Students can create their own by using more than two variables (see slide 6), by making up their own expressions or by using more than three expressions.

The final suggestion might lead students to consider the inequality in the original prompt and attempt to find values of a and b that satisfy each of the 24 permutations.

It is possible to find values of a and b that satisfy the permutations shaded in green below (1 is the lowest value and 4 the highest). There are no values of a and b that satisfy the permutations in red. That is because ab can never be greater than both a2 and b2 as would be required in those permutations.

February 2021

# An alternative prompt

In Mathematics Teaching 223 (July 2011), Geoff Tennant describes using this prompt with a class on his subject knowledge enhancement course (pre-teacher training) at Reading University.

In order to keep the arithmetic simple, the students decided to substitute perfect squares into the inequality. This led to the following results in the table:

Geoff was taken aback. A quick check of a = 1 and b = 9 led to the conjecture: If a and b are consecutive perfect squares, then ½(a + b) - √(ab) = ½. The class went on to prove the result, using a = n2 and b = (n + 1)2.