Teach Secondary Issue 13.4

[ M A T H S P R O B L E M ] Colin Foster (@colinfoster77) is a Reader in Mathematics Education in the Department of Mathematics Education at Loughborough University. He has written many books and articles for mathematics teachers. foster77.co.uk , blog.foster77.co.uk THE DIFFICULTY Usha is solving the inequality 11 − 3 > 5 . She writes: 11 − 3 > 5 − 3 > −6 > 2 How can you tell that her answer is wrong ? Where did she go wrong? Students may try substituting a value that satisfies > 2 to see whether it also satisfies the starting inequality. In fact, every value that satisfies > 2 fails to satisfy 11 − 3 > 5 . For example, if = 10 , which is greater than 2 , then 11 − 3 × 10 −19 , which is not greater than 5 . Students may struggle to explain where Usha has gone wrong, as her steps seem plausible. They might find, by substituting, that her second line is correct, but her third line is incorrect. If they say ‘ Because the inequality sign switches around when you divide by a negative number ’ challenge them to explain why . THE SOLUTION What Usha does would be correct for an equality : 11 − 3 = 5 − 3 = −6 = 2 The solution 2 does satisfy 11 − 3 = 5 , because 11 − 3 × 2 5 . But it does not work for an inequality . In this lesson, students make sense of when and why the inequality sign turns around How else could you solve this equation ? An alternative solution method is: 11 − 3 = 5 11 5 + 3 6 3 2 Try this method with the original inequality . This leads to the correct solution: 11 − 3 > 5 11 5 + 3 6 3 2 Why does this method work and Usha’s method doesn’t? The second method never multiplies or divides both sides by a negative number . Usha’s method does, because she divides by −3 . Dividing or multiplying by negative numbers changes the direction of an inequality . Students could experiment with a (true) inequality, such as 14 10 ; dividing both sides by −2 produces −7 −5 , which is false. A further challenge could be to solve this inequality, both correctly and incorrectly, and to explain the difference: 1– 1 2– Taking reciprocals of both sides, to obtain > 2 , is invalid, since, for example, 1 3 1 2 is false. Multiplying both sides by 2 , to obtain 2 is valid only if > 0 . However, we know that if a fraction is greater than 1 2 then it must be positive, in which case the numerator and the denominator must have the same sign, so 2 is correct. Checking for understanding Make up three inequalities that could catch someone out! Solve each one correctly and incorrectly, explaining why the incorrect solutions are incorrect. INEQUALITY SIGNS TURNING ROUND When solving inequalities, students are often confused when the inequality sign changes direction, says Colin Foster 21 teachwire.net/secondary