Square roots of negative numbers, and the surd rule that stops working

What to assume they know

• An equation with no real solution, and the symbol invented to solve it — the symbol i and the relation i² = −1
• Multiplying, then inverting: how division becomes possible — multiplication of numbers in the form a + ib
• The meaning of the radical sign for a positive number, and that the two square roots of a positive number differ only in sign
• What a proof by contradiction establishes, and what it does not

What they should be able to do

• List both square roots of −1 and verify each by squaring
• Explain why the usual choice of "the positive root" is unavailable for a negative number
• State the convention the chapter adopts for a radical over a negative number
• Write the square roots of a given negative number using that convention, and check them by squaring
• State the range of numbers over which the product-of-radicals rule holds, and the case in which it does not
• Reproduce the contradiction argument that rules out the both-negative case
• Evaluate a product containing a radical over a negative number by converting to the form a + ib first

Where it usually goes wrong

• "The radical over −4 times the radical over −9 is the radical over 36, so 6." This is the error the section exists to kill. Verified: the correct value is (2i)(3i), which is 6i², that is −6. Show both routes side by side and let the sign disagreement do the work.
• "So the product rule for radicals is simply false now." It is not. It holds when both numbers are positive, when exactly one is negative, and when either is zero. Precisely one case fails. Teaching it as "the rule is dead" makes students distrust valid steps.
• "The contradiction shows that i is inconsistent." It shows that the rule, applied where it was never established, is what fails. The chapter's argument begins by assuming the rule and ends by discarding it, not by discarding i.
• "The radical over −1 is ±i." The radical names one value by convention. The equation whose unknown squares to −1 has two solutions. Those are different statements and the page makes both.
• "Because i is a square root, it must be positive or negative." Neither. Nothing in this chapter compares two complex numbers as larger and smaller, so the tie-break that defined the radical over a positive number has nothing to work with here.
• "Multiply first, convert later." Reverse it. Every radical over a negative number should be turned into the form a + ib before any multiplication, which is exactly what Example 4 does in its first line.

Questions to check understanding

• Write the square roots of a given negative number and verify one by squaring
• Simplify a product of two radicals over negative numbers, where the naive rule gives the wrong sign
• Express a product such as the one in Example 4 in the form a + ib
• State the condition under which the product-of-radicals rule may be used
• Explain, in two or three lines, why the rule fails when both numbers are negative
• Items where a radical over a negative number appears inside a longer expression and must be converted before anything else is done

Examples worth working on the board

Items marked verified are worked out here from the chapter's stated data.

• Both roots of −1 (§4.3.6, p. 79). The candidates are i and −i. Verified: squaring i gives −1 by definition; squaring −i gives the same, because the two minus signs multiply away. So there are two of them, and nothing in the chapter ranks one above the other.
• The convention (§4.3.6, p. 79). The radical over −1 is taken to name i alone. Note: this is a decision about notation. The equation whose unknown squares to −1 still has two solutions, and the chapter says so on the same page.
• Both roots of −3 (§4.3.6, p. 79). The candidates are √3·i and −√3·i. Verified: squaring √3·i gives 3 times i², which is −3; the other candidate squares to the same value. The chapter then fixes the radical over −3 to mean √3·i.
• The general form (§4.3.6, p. 79). For a positive real a, the radical over −a is √a·i.
• The rule and its range (§4.3.6, p. 79). The product-of-radicals rule was established for two positive numbers. The chapter states that it survives when one of the two is positive and the other negative, and then asks what happens when both are negative.
• The contradiction (§4.3.6, p. 80). Inputs: apply the rule to the radical over −1 multiplied by itself. Verified as a valid reductio: the left side is i·i, which is i², that is −1; the right side, if the rule were allowed, would be the radical over the product of −1 with −1, that is the radical over 1, which is 1. So the assumption forces −1 to equal 1. The assumption is therefore false in the both-negative case, and the chapter says exactly that.
• The zero case (§4.3.6, p. 80). If either number is zero the two sides agree at 0. Verified trivially, and worth one line so the exception is stated completely rather than as "both negative fails, everything else is fine".
• Example 4 (p. 81). Inputs exactly as printed: the product of (−√3 + √(−2)) with (2√3 − i). Verified: convert first — the radical over −2 is √2·i, so the left factor is −√3 + √2·i. Expanding gives −2·3 in the first slot from the leading term, √3·i and 2√6·i from the cross terms, and −√2·i² which is +√2. Collecting: the first slot is −6 + √2 and the second is √3 + 2√6, which factors as √3(1 + 2√2). The chapter reaches the same result. The lesson is the order of operations: convert every radical to the a + ib form before multiplying anything.
• The Historical Note's numbers (pp. 87–88). Cardan's pair, 5 plus and minus the radical over −15, is exactly this convention in use. Verified: the radical over −15 is √15·i, whose square is −15, so the product of the pair is 25 + 15 = 40 and their sum is 10 — the two conditions the note states.

Figures to have open

• A domain map for the product rule: a two-by-two panel over the signs of the two numbers, three cells ticked and one crossed, with the zero case noted at the edges. Standard schematic, and the image this topic most needs; the chapter has no figure in §4.3.6.
• A two-route panel for the −4 and −9 example, both routes computed to the end so the two answers stand together and differ in sign. Standard schematic; the numbers are added here, not the chapter's.
• A step panel for the contradiction with the assumption tagged at the top and struck through at the bottom. Standard schematic.
• No textbook figure is required for this topic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 4, whose printed title is Complex Numbers and Quadratic Equations; §4.3.6, pp. 79–80.
• Example 4, p. 81, for a product in which the conversion has to come first.
• Historical Note, pp. 87–88, for Cardan's pair and for the older statements by Mahavira and Bhaskara that a negative quantity has no root of this kind.
• Backward pointer inside the same chapter: the multiplication rule used in Example 4 is §4.3.3 on p. 78.