Deciding parity without computing the answer

What to assume they know

• Even and odd numbers, including for negative integers
• Adding and subtracting integers, and removing a bracket that a minus sign stands in front of (When a sum of consecutive numbers is a multiple of something uses the same manipulation)
• Letter-numbers standing for any integer, and simplifying an expression by collecting like terms
• Taking a common factor out of a sum, and reading the result as "2 times something"
• Squaring a number, and the meaning of a cube written as a small raised 3
• The positive-and-negative token picture of integers from the Class 7 chapter on Integers, which this chapter refers back to by name

What they should be able to do

• List all eight ways of placing plus and minus signs between four numbers, using a branching diagram rather than trial and error
• Show that switching one sign changes the value by twice one of the terms, and conclude that all eight values share one parity
• State the three parity rules for a sum or difference and apply them along a chain of terms
• Decide whether an arithmetic expression is even without evaluating it
• Decide whether an algebraic expression is even for every integer value of its letters, and give an example and a non-example where it is not
• Justify "always even" two ways — by parity of the parts, and by exhibiting 2 as a factor — and say what each justification does that the other does not
• Recognise that a claim quantified over all integers cannot be settled by substituting numbers, but can be destroyed by one substitution

Where it usually goes wrong

• "Negative numbers are neither even nor odd." The chapter takes the trouble to say otherwise on Part I p.113, and it must, because half the printed results in the sign-switching activity are negative. An even number is one with 2 as a factor; the sign is irrelevant.
• "Checking a few cases shows it always happens." Part I p.115 says outright that the supply of four-number choices is unlimited. Eight expressions for one run is a demonstration; the sign-switch argument is a proof. Keep the two visibly apart.
• "Switching a sign changes parity, because minus is different from plus." It changes the total by twice a term. Twice anything is even, and shifting by an even amount cannot cross between even and odd. This is the load-bearing step of the whole topic.
• "3g + 5h is even because 3 + 5 is even." Coefficients do not add like that. The parity of each term depends on its letter, and odd × odd stays odd.
• "x² + 2 is even because 2 is even." The square carries the parity of x, so the expression is even only for even x. The chapter's 38-and-11 pair is there precisely to break this.
• "13k − 5k is odd because 13 and 5 are odd." Simplify first: it is 8k. Students who read parity off unsimplified coefficients get this backwards.
• "Showing 2 is a factor and showing both parts are even are the same argument." They are not, and the chapter gives both for a reason. The factor argument survives when the parts are not separately even — 6m + 2n is 2(3m + n) whether or not 3m is even — while the parts argument does not.
• "One value makes it true, so it is always true." One value can only refute. The example / non-example pairing on Part I p.116 is a lesson in what each kind of instance is worth.

Questions to check understanding

• List every expression obtainable by signing a given set of numbers, and state the count before listing
• Given one expression and its value, state the parity of every other expression from the same set without computing
• Decide the parity of an arithmetic expression built from large numbers, and justify it from the parity of the parts
• Decide whether a given algebraic expression is even for every integer value of its letters, and supply an example and a non-example when it is not
• Rewrite an expression to display a factor of 2, and say what that display proves
• Write your own expression that is even for every integer value, and one that is even only sometimes
• Explain why testing values cannot establish an "always" claim, using this chapter's own reasoning

Examples worth working on the board

Inputs. The chapter prints the values marked "on the page"; the rest are left blank for the student.

• The branching diagram (Part I, §5.1, p.113). A tree rooted at 3, branching to 4 twice (once under a plus, once under a minus), each 4 branching to 5 twice, each 5 branching to 6 twice — eight leaves. Two dashed arrows lead out from the top two leaves to their expressions, set in red: the all-plus one, and the one that differs only in the last sign. The other six expressions are for the student to write. Note the structure: three gaps, two choices per gap, so 2 × 2 × 2 leaves.
• The evaluated expressions, printed on the page (Part I, §5.1, p.113, three dotted boxes). First box, run 3, 4, 5, 6: 3 + 4 − 5 + 6 = 8 and 3 − 4 − 5 − 6 = −12. Second box, run 5, 6, 7, 8: 5 + 6 − 7 + 8 = 12 and 5 − 6 − 7 − 8 = −16. Third box is entirely blank, for a run the class chooses. All four printed values are even, two of them negative — which is exactly why the page pauses to say that −2, −4 and −6 are even too.
• Explanation 1 (Part I, §5.1, p.114). Worked on the page: start from a + b − c − d, replace the plus before b by a minus, and subtract the new value from the old. The bracket expansion is set out line by line and the difference is 2b. The page then turns the question round — flip a minus to a plus instead — and hands that case to the student.
• Explanation 2 (Part I, §5.1, p.114). Three parity rules printed with a plus-or-minus sign, so each covers a sum and a difference at once: odd with odd gives even, even with even gives even, odd with even gives odd. The page then chains them across a ± b, then a ± b ± c, then a ± b ± c ± d.
• Explanation 3 (Part I, §5.1, p.115). The same eight-leaf tree redrawn with the letters a, b, c, d in place of 3, 4, 5, 6, set to the right of a paragraph that points back to the token model from the Integers chapter and invites the student to reconstruct the argument with tokens. The page states that the ways of choosing four numbers are unlimited, which is the sentence that makes the algebra necessary rather than decorative.
• "Breaking Even", the arithmetic set (Part I, §5.1, p.115, eight tinted boxes). 43 + 37; 672 − 348; 4 × 347 × 3; 708 − 477; 809 + 214; 119 × 303; 543 − 479; and 513 cubed. The instruction is to decide evenness without computing. Note that the set deliberately mixes sums, differences, a product of three factors and a power, so the parity rules for multiplication have to be reasoned out rather than recalled.
• "Breaking Even", the algebraic set (Part I, §5.1, p.115, nine tinted boxes). Four of them are built from two letters with a coefficient each: 2a + 2b, then 3g + 5h, then 4m + 2n, and 2u − 4v. One more puts the same letter in both terms: 13k − 5k. Another, 6m − 3n, carries an even coefficient in front but a factor of 3 that leaves the parity to n. Two carry a square: x² + 2, and b² + 1. The last is a product rather than a sum, 4k × 3j. The question is which are even for every integer value of the letters. Two are traps worth planning around: 13k − 5k wears two odd coefficients and collapses to a multiple of 8, and 6m − 3n has an even coefficient in front but a factor of 3 that leaves the parity to n.
• The fully worked case (Part I, §5.1, p.116). The chapter works 4m + 2q two ways: as an even plus an even, and as 2 × (2m + q), which shows 2 as a factor. Printed numerical check: m = 4 and q = −9 give −2. Note the letter change — the tinted box on the previous page reads 4m + 2n and the worked paragraph uses q. Same expression, different second letter.
• The negative case (Part I, §5.1, p.116). For x² + 2 the chapter gives an example and a non-example: x = 6 gives 38, x = 3 gives 11. Hand both to the explanation — the pair is the point, not either one alone.
• Closing prompt (Part I, §5.1, p.116): write expressions of your own that are even for every integer value. Good exit task.

Figures to have open

• The eight-leaf branching diagram, drawn twice: once with 3, 4, 5, 6 and once with a, b, c, d. Both are the chapter's own figures (Part I, §5.1, Part I pp.113 and 115) and the topic is hard to carry without them. Redraw as a clean schematic rather than reproducing the printed art.
• A token strip: a row of positive counters and a row of negative counters pairing off, with the unpaired remainder highlighted. Standard schematic. The chapter names this model and points to it but does not draw it here.
• A two-column sorting board for the nine algebraic expressions, with room for an example and a non-example beneath any expression placed in the right-hand column. Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part I, printed Chapter 5, "Number Play", §5.1 "Is This a Multiple Of?", Part I pp.113–116. Within §5.1 the relevant printed subheading is "Breaking Even" (Part I p.115); the sign-switching material before it runs on from the opening subheading with no heading of its own.
• The three explanations are labelled on the page as Explanation 1 (Part I p.114), Explanation 2 (Part I p.114) and Explanation 3 (Part I p.115); they can be cited by those labels.
• Part I p.115 carries two owl-illustrated boxes of advice — one on the value of hearing how others solved a problem, one on conjecturing as part of doing mathematics. Section 9 is the natural place to use the second.
• Backward pointer: the token model belongs to the Class 7 chapter on Integers, which this page names.