Why a negative times a negative must be positive

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What to assume they know

Negative numbers on the number line, and adding and subtracting them: the token model, the zero pair, additive inverse, and that subtracting is adding an inverse
Multiplication of whole numbers read as repeated addition, and the times tables
Class 6: adding and subtracting integers
Reading a descending pattern and predicting its next term
Factorise and regroup instead of multiplying head-on: rearranging a product without changing it

What they should be able to do

Interpret 4 × 2 and 4 × (−2) as placing tokens into an empty bag, and state what the multiplier and the multiplicand each control
Name the multiplier, the multiplicand and the product in a written multiplication, using this chapter's convention
Explain why a negative multiplier is modelled by removal rather than placing
Carry out (−4) × 2 and (−4) × (−2) with tokens, inserting zero pairs first, and say why the bag must start empty
Re-express a removal of positives as an addition of negatives, and get the same product
Continue a descending multiplication ladder past zero and state the constant step it keeps
State the four sign outcomes of multiplying two integers, and attribute each to the evidence that produced it
Use 1 × a = a and −1 × a = −a for any integer a, and connect the second to the additive inverse

Where it usually goes wrong

"Multiplying always makes things bigger." 4 × (−2) is further from zero but smaller than 4; (−4) × (−2) is positive but nothing was ever positive to start with. Treat "bigger" and "further from zero" as two different questions from the first section.
"A negative times a negative is positive because two minuses cancel, like in language." The chapter never argues from grammar, and the grammar analogy fails immediately for a negative plus a negative. The reason is the token model plus the unbroken ladder — say the reason, not the mnemonic.
"(−4) × 2 must be worked out differently from 4 × (−2), since the roles differ." They are modelled differently on pp.29–30 — one places, one removes — and they still land on the same value. That coincidence is worth pausing on; it is the seed of the commutativity argument in Commutative, associative, and distributive over the integers.
"You can only remove what is already in the bag." Every new operation starts empty on purpose. Inserting zero pairs first is what makes removal always available, exactly as it did for subtraction on p.28.
"The answer depends on how you draw the tokens." The p.31 task exists to kill this. Three different-looking sets all stand for −2 and all give the same answer when placed four times.
"The pattern proves the rule." The ladders are strong evidence, not a deduction from nothing — what they actually show is that keeping the step constant forces these values if the extension is to be consistent. Present them as the chapter does: a reason to accept the extension, alongside the token argument, not instead of it.
"Since the product's magnitude never changes, the signs must not matter." The magnitude is indeed decided by the magnitudes alone. The sign is decided separately. Splitting a product into those two independent questions is the cleanest thing a student can take from p.33.

Questions to check understanding

Find a product with tokens and describe the placing or removal in words (the chapter's own task, four items, p.31)
Given one product of two large numbers, write down the three sign variants without recalculating (p.31)
State the sign of a product from the signs of its two factors, in all four cases
Continue a descending multiplication ladder past zero and state its constant step
Fill a blank factor so a multiplication statement comes out true (p.39)
Given a number, name the integer that multiplies with −1 to give it (p.42)
Explain, without using the word "rule", why removing negatives from a bag leaves it positive
Word problems that turn a rate and a count into a product with a sign — marks per answer, degrees per hour, rupees per bag — are the standard competency dress for this content

Examples worth working on the board

4 × 2 with tokens (Part II, §2.2, p.29). Checked against the printed page. Four rounded boxes, each holding two green tokens, with a brace beneath. Input: two greens, placed four times.
4 × (−2) with tokens (Part II, §2.2, p.29). Checked against the printed page. The same layout in red. Beneath it the printed statement carries three callout bubbles pointing at the three parts of the equation — the first factor, the second factor, the answer. That callout is the source for section 3. The three words in the bubbles do come through in a text extraction, but the leader lines that say which part of the equation each bubble points at do not, so it is the printed page, not the extract, that settles this chapter's convention.
(−4) × 2 with tokens (Part II, §2.2, p.30). Checked against the printed page. Four boxes, each holding two zero pairs, with the green token in each pair struck through by a diagonal line. Inputs: two positives to be removed, four times, starting from nothing.
(−4) × (−2) with tokens (Part II, §2.2, p.30). Checked against the printed page. The same four boxes of zero pairs, this time with the red tokens struck through. Inputs as above with the colour swapped.
The four results the token model establishes (Part II, §2.2, p.30). Printed as a stacked list: 4 × 2, 4 × (−2), (−4) × 2, (−4) × (−2), all built from the same pair of magnitudes.
Three drawings of the same number (Part II, "Figure it Out", p.31). Checked against the printed page. (a) two red tokens. (b) two reds on an upper row, and below them two reds and two greens. (c) four reds on an upper row, and below them two reds and four greens. Each set is worth −2. The task is to place each set four times and see whether the answer depends on which drawing was used. This is the chapter's own well-definedness check and it deserves its own beat.
Removal rewritten as addition (Part II, pp.31–32). Checked against the printed page. Beside the text sits a column of eight red tokens labelled (−4) × 2. Input: removing two greens has the same effect as adding two reds, so the same product arrives with nothing ever taken out.
The three ladders (Part II, "Patterns in Integer Multiplication", pp.32–33). Checked against the printed page; the step labels sit in curved arrows beside the equations, and although the labels themselves extract they come out in a block of their own, detached from the rows they belong to. Ladder A: 4 × 3 down to 0 × 3, arrows marked −3. Ladder B: the same continued to (−1) × 3, (−2) × 3, (−3) × 3, arrows still marked −3. Ladder C: 4 × (−3) down to 0 × (−3) and on to (−1) × (−3), (−2) × (−3), (−3) × (−3), arrows marked +3 throughout.
The four times-3 columns (Part II, p.33). Checked against the printed page. A boxed four-column table, each column running from 1 to 10 in the multiplier: column 1 is n × 3, column 2 is (−n) × 3, column 3 is n × (−3), column 4 is (−n) × (−3). Four columns of ten: forty products, four sign combinations, one magnitude pattern. This is the single most useful figure in the topic.
1 × a and −1 × a (Part II, pp.33–34). Inputs: placing a into an empty bag exactly once leaves a, whatever the sign of a; and multiplying by −1 gives something with the same magnitude and the opposite sign. The chapter's own conclusion names the result the additive inverse.
Products from a known product (Part II, "Figure it Out", p.31). Input: 123 × 456 = 56088. Three sign variants are then asked for without recalculating.

Figures to have open

Green and red circular tokens carrying + and −, groupable into small boxes, with a struck-through state for removal. Standard schematic; this is the chapter's central visual and it must be redrawn rather than lifted.
A bag or container that can be shown empty and then filled, for sections 1–6.
A ladder layout: a column of equations with a curved arrow between consecutive rows carrying a step label. Must support both a −3 and a +3 label without redrawing.
A four-column times table, ten rows deep, that can highlight one column, one row, or the magnitude of every entry at once.
The three-part callout on 4 × (−2) = −8, redrawn from the printed page. The three labels survive text extraction; what is lost is which part of the equation each one points at, which is the whole content of the figure.
No photograph or dataset from the textbook is needed.

Where this sits in the book

NCERT Class 7 Mathematics, Ganita Prakash Part II, printed Chapter 2 "Operations with Integers", §2.2 "Multiplication of Integers", pp.29–30; the "Figure it Out" block on pp.31–32; the unnumbered sub-heading "Patterns in Integer Multiplication", pp.32–33; and the opening of the following "Figure it Out" block, pp.33–34, for 1 × a and −1 × a.
Part II, p.45, SUMMARY, for the chapter's own closing statement of the sign cases and the four-quadrant table that carries it.
Backward pointer: Negative numbers on the number line, and adding and subtracting them for the token model, the zero pair and the additive inverse this section reuses without re-deriving.