Re-running the same contradiction on other prime square roots

What to assume they know

• Assume the square root of 2 is a fraction, and watch the assumption collapse — the contradiction argument for the square root of 2, in full
• A prime dividing a square must divide the number itself — Theorem 1.2, and that its hypothesis of primality cannot be dropped
• That a difference of two integers is an integer, and that an integer over a non-zero integer is rational
• Rearranging a linear equation to isolate one term, including moving a multiplier across

What they should be able to do

• Extract the template from the square-root-of-2 proof: name every place the number 2 entered, and replace it with a slot
• Instantiate the template at 3 and at 5, and check that the two applications of Theorem 1.2 remain legitimate
• State the chapter's general claim about square roots of primes, and say exactly which step of the template forces the restriction to primes
• Show that the template stalls at 4, and resist the false generalisation that roots of composites are therefore rational
• Prove a result of the form "a rational minus an irrational root" by isolating the root, and name the closure property used
• Prove a result of the form "a rational times an irrational root" the same way, using division instead of subtraction
• Recognise which of two given irrationality questions needs the full contradiction machinery and which needs only one rearrangement

Where it usually goes wrong

• "Each of these is a separate proof to memorise." There is one proof, run at different primes, and then one rearrangement, run on different expressions. Naming which of the two a question needs is most of the work.
• "The argument shows any square root is irrational." It stalls at 4 in a visible place, and 4 is not an exotic exception — every perfect square behaves the same way. The chapter's general claim is stated over primes for exactly this reason.
• "So the square root of a composite number is rational." That does not follow either, and it is the more damaging error of the two. The stall at 4 shows the template does not settle the composite case, not that the case comes out the other way. Say so directly, and leave the composite case open.
• "Example 6 needs the whole contradiction machinery again." It needs one line of rearranging plus a result that finished a few lines above it on the same page. Students who restart from coprime integers have not seen what the example is for.
• "A rational plus an irrational could be rational." If it were, subtracting the rational back off would leave a rational, and the irrational would be rational. That single sentence is the whole content of the first Class IX fact, and it is worth saying rather than asserting the fact.
• "3 times a root is irrational because 3 is not a root." The reason is that a rational divided by a non-zero rational is still rational.

Questions to check understanding

• Prove that the square root of a stated prime is irrational
• Prove irrationality for an expression built by adding, or by subtracting, a whole number and a square root
• Prove that a stated multiple or reciprocal of an irrational root is irrational
• Given a question, say whether it needs the contradiction argument or only a rearrangement, with the reason
• Explain why the standard argument does not establish that the square root of a perfect square is irrational

Examples worth working on the board

Values marked verified are worked out here of the exercises; the chapter prints no answers, and no answer key was consulted.

• The template, with the slot marked (§1.3, pp. 6–8). The lines that carry a number are: the assumed equality; the multiplied-up form; the squared form, where the prime multiplies the denominator's square; the first appeal to Theorem 1.2; the substitution of the numerator as the prime times a new integer; the simplified equation; and the second appeal to Theorem 1.2. Seven places, one number. Show the template with the slot empty before filling it.
• Example 5 — the square root of 3 (§1.3, pp. 7–8). Verified: three times the denominator's square equals the numerator's square; 3 divides the numerator's square so 3 divides the numerator; writing the numerator as 3c gives three times the denominator's square equal to 9c², so the denominator's square is 3c², so 3 divides the denominator. Both parts carry a factor of 3, against the coprime assumption.
• Exercise 1.2 Q1 — the square root of 5 (§1.3, p. 9). Input only; the student fills the slot. Verified: the two intermediate equations are five times the denominator's square equal to the numerator's square, and, after substituting 5c, the denominator's square equal to 5c². Theorem 1.2 is used with p = 5 at both appeals.
• The stall at 4 (not in the book). Verified: four times the denominator's square equals the numerator's square; 2 divides the numerator, so it is 2c; substituting and dividing by 4 leaves the denominator's square equal to c² with no prime left over, and the argument halts. It has to: the square root of 4 is 2, which is a perfectly good fraction. Contrast this with 3 and 5, where a factor survives the substitution every time.
• Example 6 — 5 minus the square root of 3 (§1.3, p. 8). Verified: if the whole expression were a fraction a over b, then the root equals 5 minus a over b, which collects into (5b − a) over b — a ratio of two integers with a non-zero bottom, so the root would be rational — and that has been ruled out ten lines higher up the same page, where Example 5 closes. No second contradiction argument is run. Numerically the value is about 3.2679, which the explanation may show but must not treat as evidence.
• Example 7 — 3 times the square root of 2 (§1.3, pp. 8–9). Verified: if the product were a over b, the root equals a over 3b, again a ratio of integers with a non-zero bottom. Value about 4.2426. Same manoeuvre, division in place of subtraction.
• The two facts recalled from Class IX (§1.3, p. 8). The chapter restates that combining a rational with an irrational by addition or subtraction leaves it irrational, and that multiplying or dividing an irrational by a rational other than 0 does the same — and is explicit that only particular cases are established here.
• Exercise 1.2 Q2 and Q3 (§1.3, p. 9). The items are: 3 plus twice the square root of 5; the reciprocal of the square root of 2; 7 times the square root of 5; and 6 plus the square root of 2. Verified isolations, which is all each one needs: for the first, the root of 5 equals half of (the assumed rational minus 3); for the second, if the reciprocal equals a over b then the root of 2 equals b over a, and a cannot be 0 because the reciprocal is not 0; for the third, the root of 5 equals the assumed rational over 7; for the fourth, the root of 2 equals the assumed rational minus 6. Verified approximate values, for the visual only: about 7.4721, 0.7071, 15.6525 and 7.4142. Note that the second item is the only one where a zero denominator has to be argued away — that is the item worth working.

Figures to have open

• A template panel with a visible slot, refillable with 3, then 5, then 4. This is an added device and carries sections 1 to 5; the chapter simply writes the argument out again.
• A dependency tree with the square root of 2 at the base, the other prime roots as siblings, and the combination results hanging off them. The chapter's structure implies this ordering but prints no such diagram.
• An isolation strip that shows a rearrangement one term at a time and ends with the root alone on one side. Standard schematic.
• No textbook figure is required; §1.3 carries no artwork on pp. 6–9.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, printed Chapter 1 "Real Numbers", §1.3 "Revisiting Irrational Numbers": the general claim about square roots of primes on p. 6; Example 5 across pp. 7–8; the two recalled Class IX statements and Example 6 on p. 8; Example 7 across pp. 8–9; Exercise 1.2 on p. 9.
• Theorem 1.2 on p. 6 and Theorem 1.3 on pp. 6–7, on which everything here depends.
• §1.4 "Summary", p. 9, which lists the irrationality proofs as one of the chapter's three outcomes.