Proofs in Mathematics (Appendix)

This appendix has five moving parts, and NCERT builds them in a strict order because each depends on the one before it.

Mathematical statements (Section A1.2) sets the ground rule: mathematics only works with sentences that are unambiguously true or unambiguously false. Orders, questions, exclamations, and sentences that depend on unstated context ("vehicles have four wheels") are excluded. You practise sorting real sentences into always-true, always-false, and ambiguous, and you learn to restate an ambiguous or false conditional so that it becomes true — for example "if the diagonals of a quadrilateral are equal, then it is a rectangle" is false as stated (a random quadrilateral with equal diagonals need not even be a parallelogram), but restated with the extra hypothesis "parallelogram" it becomes a true theorem.

Deductive reasoning (Section A1.3) is the engine that turns premises into conclusions. Given "all mathematics textbooks are interesting" and "you are reading a mathematics textbook," deduction hands you "you are reading an interesting textbook" — and this chain of logic is valid whether or not the premise about textbooks is actually true. This section also introduces two named proof strategies used throughout school geometry: the direct method, where each statement follows from the last, and proof by exhaustion, where you list every possible case and eliminate all but one.

Negation (Section A1.5) teaches you to write "$\sim p$," the statement that is true exactly when $p$ is false, and warns you about the single commonest trap: negating "All teachers are female" is NOT "All teachers are not female" (which absurdly claims all teachers are male) — it is "Some teacher is not female."

Converse (Section A1.6) takes a compound statement of the form "if $p$ then $q$" (written $p \Rightarrow q$) and swaps the two halves to get "if $q$ then $p$" (written $q \Rightarrow p$). The chapter's central warning is that a true statement's converse need not be true — "if $n$ is even then $2n+1$ is odd" is true, but its converse "if $2n+1$ is odd then $n$ is even" is false: take $n = 7$, an odd integer; then $2n + 1 = 15$ is odd, satisfying the converse's hypothesis, yet $n = 7$ itself is not even, so the converse's conclusion fails.

Proof by contradiction (Section A1.7) is the capstone: assume the negation of the claim, deduce validly, arrive at a contradiction (a statement and its own negation both appearing true), and conclude that the original claim must hold. This is the exact method Class 9 and Chapter 1 used to prove $\sqrt 2$ is irrational, and this chapter both reviews that proof and adds new ones — the converse of the Pythagoras theorem, and a geometric minimum-distance result. Two worked derivations below carry this method through completely: one proves that every prime greater than $3$ has the form $6k+1$ or $6k+5$, first directly by exhaustion and then again by contradiction; the other proves the converse of the Pythagoras theorem by contradiction and construction, exactly as NCERT does.

What counts as a mathematical statement

A sentence qualifies as a mathematical statement only if it is unambiguously always true or always false — never a matter of interpretation, mood, or missing context. Orders ("finish your homework"), questions ("which teams are in the final?"), and exclamations ("what a goal!") are automatically excluded because they carry no truth value at all. Among ordinary declarative sentences, some are still excluded for being ambiguous: "vehicles have four wheels" depends entirely on which vehicle you mean, so it is neither always true nor always false — it is not a mathematical statement. This filtering matters because every later tool in this chapter — negation, converse, deduction, contradiction — is defined in terms of the truth or falsity of a statement. NCERT's working test is simple: is there a single, context-free answer of "true" or "false" that everyone competent would agree on? "The speed of light is approximately $3 \times 10^5$ km/s" passes (always true, by measurement). "A road to Kolkata will be closed from November to March" fails — which road? Quantified statements need special care. "All equilateral triangles are isosceles" is a statement about every member of a class, and it is true — but "all isosceles triangles are equilateral" is false, since a triangle with sides $5, 5, 8$ is isosceles but not equilateral; that one counter-example is enough to kill the "all" claim. "Some rational numbers are integers" and "some rational numbers are not integers" are both true simultaneously — "some" only claims existence of at least one example, and both kinds of rational number exist.

Deductive reasoning: premises to conclusion

Deductive reasoning is the process of arriving at a conclusion that is forced by a set of given statements, called premises or hypotheses, using only valid logical steps — never by guessing, measuring, or checking a few cases. The crucial and slightly uncomfortable feature of deduction is that it does not care whether the premises are true in the real world; it only guarantees that if the premises are true, the conclusion follows. Given "Bijapur is in Karnataka" and "Shabana lives in Bijapur," deduction hands you "Shabana lives in Karnataka" — a rock-solid conclusion built from two assumed facts, without anyone needing to verify a map. This distinction — validity of reasoning versus truth of premises — is the chapter's most important abstract idea. NCERT makes it vivid with the prime number $19423$: given the hypothesis "$\sqrt p$ is irrational for every prime $p$" and the (unverified) hypothesis "$19423$ is prime," deduction concludes "$\sqrt{19423}$ is irrational" — a perfectly valid conclusion drawn without anyone actually checking that $19423$ is prime. If that unverified premise later turns out false, the conclusion could be wrong too, but the reasoning process itself was not at fault — every theorem is stated as "if the hypotheses hold, then the conclusion holds," and a proof only certifies the implication. Two named strategies for building a full proof, direct and exhaustive, sit inside deductive reasoning: The direct (deductive) method builds a numbered chain of statements, each justified by a definition, an axiom, an earlier theorem, or a valid step from the line before it, ending at the statement you set out to prove. Example 10 of the chapter proves "the sum of two rational numbers is rational" this way: write $x = m/n$ and $y = p/q$ with $n, q \neq 0$, compute $x + y = (mq + np)/(nq)$, note that $mq + np$ and $nq$ are both integers with $nq \neq 0$ (properties of integers), and conclude $x + y$ fits the definition of a rational number.

Proof by exhaustion lists every case a variable can fall into and eliminates all but the wanted one. To show every prime $p > 3$ has the form $6k+1$ or $6k+5$: Euclid's division lemma says dividing $p$ by $6$ leaves one of six remainders, so $p$ is $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$ for some integer $k$. But $6k, 6k+2, 6k+4$ are all even (divisible by $2$) and $6k+3$ is divisible by $3$, so none of these four can be prime once $p > 3$. Only $6k+1$ and $6k+5$ survive the sweep. This method is worked in full as the first derivation below.

Negation: writing "it is false that..." correctly

The negation of a statement $p$, written $\sim p$ (read "not $p$"), is the statement that is true exactly when $p$ is false, and false exactly when $p$ is true. This defining rule — $\sim p$ and $p$ never share a truth value — is the only test you need to check whether a candidate sentence is really the negation of another. For a simple statement, negation is often just prefixing "it is false that": if $s$ is "$2 + 2 = 4$," then $\sim s$ is "$2 + 2 \neq 4$." Applying negation twice returns the original statement: $\sim(\sim p)$ is $p$, for any statement $p$. The genuine difficulty appears with quantified statements built from "all" or "some." Consider $q$: "All teachers are female." Naively inserting "not" gives "All teachers are not female" — but that sentence claims every single teacher is male, a much stronger and different claim than simply denying $q$. The correct negation of "All A are B" is "Some A are not B." Similarly the negation of "Some A are B" is "No A are B." NCERT's working rule: write the statement with "not" first, then check for confusion around "all"/"some" and rephrase if the naive insertion misleads.

Converse: swapping hypothesis and conclusion

A compound statement built from two simple statements $p$ and $q$ using "if...then" is written $p \Rightarrow q$ (read "$p$ implies $q$"), where $p$ is called the hypothesis and $q$ the conclusion. The converse of $p \Rightarrow q$ is obtained by simply interchanging hypothesis and conclusion, giving $q \Rightarrow p$. Note that $p \Rightarrow q$ and $q \Rightarrow p$ are converses of each other — the relation is symmetric, not one-directional. Writing a converse is a purely mechanical, syntactic operation — it says nothing about whether the resulting statement is true. This is the chapter's second big warning, parallel to the negation warning: a statement can be true while its converse is false. "If Ahmad is in Mumbai, then he is in India" is true, but its converse "If Ahmad is in India, then he is in Mumbai" is false — he could be anywhere else in the country. In geometry the same asymmetry shows up constantly: "if a quadrilateral is a parallelogram, then each pair of opposite sides is equal" is a proved theorem, and its converse ("if each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram") also happens to be true — but that is never automatic; each direction needs its own proof or its own counter-example.

Proof by contradiction: assume the opposite, force a clash

A contradiction, in the technical sense used here, occurs when a chain of valid reasoning forces both a statement $p$ and its negation $\sim p$ to be true at the same time — an outcome that logic forbids absolutely. Proof by contradiction exploits this impossibility as a proof technique: to prove that a statement $p$ is true, assume instead that $\sim p$ is true, reason validly from that assumption (using only definitions, axioms, and previously proved results), and if the chain of reasoning produces a contradiction, the fault can only lie in the starting assumption — so $\sim p$ must be false, which means $p$ is true. The method has a fixed shape worth memorising as a checklist: (1) state clearly what you want to prove, $p$; (2) assume $\sim p$ instead; (3) deduce consequences of $\sim p$ using only valid, justified steps; (4) arrive at a statement that contradicts a known fact, an earlier step, or the hypothesis itself; (5) conclude the assumption was wrong, so $p$ holds. This is exactly the machinery you already used in Chapter 1 to prove $\sqrt 2$ is irrational: assume $\sqrt 2 = a/b$ in lowest terms (assume the negation — that $\sqrt 2$ is rational), deduce that $a$ and $b$ must both be even, contradicting "lowest terms," and conclude $\sqrt 2$ is irrational after all. Two full worked proofs by contradiction — the converse of the Pythagoras theorem, and the "prime greater than 3" result via a second route — appear as the derivations below.

Board level - restating a false conditional so it becomes true

Restate the statement "If the diagonals of a quadrilateral are equal, then it is a rectangle" with an appropriate extra condition so that it becomes a true statement, and briefly justify why the original, unrestated version is false.

Step 1. Test the original statement for a counter-example. Take an isosceles trapezium: it can have two diagonals of equal length while clearly not being a rectangle (its two pairs of opposite sides are not both equal, and its angles are not all right angles). So the plain statement "if the diagonals of a quadrilateral are equal, then it is a rectangle" is false as it stands — one counter-example is sufficient to establish this.

Step 2. Identify what extra hypothesis rules out the counter-example. The trouble is that "quadrilateral" is too broad a starting class; restricting to "parallelogram" removes shapes like the isosceles trapezium, since a parallelogram already has both pairs of opposite sides equal and parallel.

Step 3. Restate: "If the diagonals of a parallelogram are equal, then it is a rectangle." This is a genuine, previously-established Class 9/10 theorem: equal diagonals in a parallelogram force all four angles to be right angles.

Step 4. Confirm the logic pattern used: NCERT's general working rule for restating a false conditional $p \Rightarrow q$ is to strengthen the hypothesis $p$ (add a condition) until the implication becomes true, without weakening the conclusion $q$. Answer: the true, restated version is "If the diagonals of a parallelogram are equal, then it is a rectangle."

Board level - negation and converse of a compound statement

Consider the statement: "If $x$ is an irrational number, then the decimal expansion of $x$ is non-terminating non-recurring." (a) Write its converse. (b) State whether the converse is true or false, with justification. (c) Write the negation of the original statement.

Step 1. Identify hypothesis and conclusion. Here $p$: "$x$ is an irrational number," and $q$: "the decimal expansion of $x$ is non-terminating non-recurring." The given statement is $p \Rightarrow q$.

Step 2. Converse. Swap hypothesis and conclusion to get $q \Rightarrow p$: "If the decimal expansion of $x$ is non-terminating non-recurring, then $x$ is an irrational number."

Step 3. Check the converse for truth. This is in fact the defining characterisation of irrational numbers taught in Chapter 1 — a real number's decimal expansion is non-terminating non-recurring if and only if the number is irrational. So the converse here is true (unlike many converses in this chapter, which fail).

Step 4. Negation of the original statement. Write $\sim(p \Rightarrow q)$ carefully: it is not simply "if $x$ is irrational then the decimal expansion terminates or recurs" (that would be a different implication). The correct negation of "if $p$ then $q$" describes a situation where $p$ holds but $q$ fails: "There exists an irrational number $x$ whose decimal expansion is not non-terminating non-recurring" — equivalently, "it is false that every irrational number has a non-terminating, non-recurring decimal expansion." Answer: the converse ("if the decimal expansion is non-terminating non-recurring, then x is irrational") is true; the negation is "there is an irrational number whose decimal expansion is not non-terminating non-recurring," which is in fact impossible to satisfy — showing that the original implication was airtight in both directions.

Higher-order thinking (HOTS) - proving by contradiction that root 3 is irrational, mirroring the chapter's method

Use proof by contradiction, following the exact five-step shape used in this chapter for the "prime greater than 3" theorem, to prove that $\sqrt 3$ is an irrational number. You may use the fact that if $3$ divides $a^2$ for an integer $a$, then $3$ divides $a$.

Step 1. State the goal clearly: we want to prove $p$: "$\sqrt 3$ is irrational."

Step 2. Assume the negation, $\sim p$: "$\sqrt 3$ is rational." By the definition of a rational number, this means $\sqrt 3 = a/b$ for some integers $a, b$ with $b \neq 0$, and we may take $a$ and $b$ to be coprime (no common factor other than $1$) — this is always possible by cancelling any common factor first.

Step 3. Deduce validly from the assumption. Squaring both sides, $3 = a^2/b^2$, so $a^2 = 3b^2$. This shows $3$ divides $a^2$. Using the given fact (if $3 \mid a^2$ then $3 \mid a$), we conclude $3$ divides $a$, so we may write $a = 3c$ for some integer $c$.

Step 4. Substitute back: $a^2 = 3b^2$ becomes $(3c)^2 = 3b^2$, i.e. $9c^2 = 3b^2$, i.e. $b^2 = 3c^2$. This shows $3$ divides $b^2$, and by the same given fact, $3$ divides $b$.

Step 5. Reach the contradiction: we have now shown $3$ divides both $a$ and $b$, meaning $3$ is a common factor of $a$ and $b$. This directly contradicts the assumption in Step 2 that $a$ and $b$ are coprime (no common factor other than $1$). Since assuming $\sim p$ led validly to a contradiction, $\sim p$ must be false, and so $p$ is true. Answer: $\sqrt 3$ is irrational, proved by exactly the same five-step contradiction shape used for the "6k+1 or 6k+5" theorem in this chapter. Timing note: about 4 to 5 minutes for a student who has already internalised the root-2 proof from Chapter 1, since the structure transfers directly — the only new piece of reasoning is quoting or re-deriving the divisibility fact "if 3 divides a squared then 3 divides a." Examiners reward explicitly labelling each of the five stages (goal, assumption, deduction, contradiction, conclusion) rather than just writing the algebra, because this chapter is specifically testing whether you recognise and can narrate the shape of the argument, not merely reproduce memorised algebra.

Students most often lose marks in this chapter not on the algebra but on the labelling of logical moves — calling something a negation when it is really a converse, or believing a plausible pattern is already a theorem.

• Treating a conjecture verified in several cases as proved. The region-counting pattern $1, 2, 4, 8$ for $n = 1,2,3,4$ points on a circle tempts the guess $2^{n-1}$, and it even survives $n = 5$ (giving $16$) — but $n = 6$ gives $31$, not $32$, killing the conjecture. No finite number of successful checks is a proof; only a general argument or an explicit counter-example settles the matter.
• Negating "all" as "all...not" instead of "some...not." The negation of "All teachers are female" is "Some teacher is not female," never "All teachers are not female" (which wrongly claims every teacher is male).
• Assuming a converse is true because the original statement is true. "If two triangles are congruent, then their corresponding angles are equal" is true; its converse ("if corresponding angles are equal, then the triangles are congruent") is false — equal angles alone permit similar triangles of different sizes.
• Confusing "restating a statement to make it true" with "finding its converse" or "finding its negation." These are three different operations on three different targets, and exam answers frequently blur them.
• Forgetting that a contradiction must be an actual logical clash, not just an unexpected or ugly-looking number. The contradiction must be a statement whose negation was already established (a definition, an earlier line of the same proof, or a stated hypothesis) — "the numbers got complicated" is not a contradiction.

• Decide whether a given sentence is a mathematical statement (always true or always false) or is ambiguous, and justify the decision.
• Restate a false or ambiguous conditional statement by adding an appropriate extra hypothesis so that it becomes true.
• Use deductive reasoning to derive a conclusion validly from given premises, independent of whether the premises are actually true.
• Identify and apply the direct method and proof by exhaustion to construct a complete, step-justified proof.
• Write the correct negation of a simple statement, and of a quantified statement containing "all" or "some," without falling into the "all...not" trap.
• Identify the hypothesis and conclusion of an "if...then" statement, write its converse, and determine independently whether that converse is true or false.
• State and apply the five-step shape of proof by contradiction: state the goal, assume its negation, deduce validly, reach a contradiction, conclude the goal is true.
• Reproduce, from the contradiction method, the proof that a given prime greater than 3 has the form 6k+1 or 6k+5, and the proof of the converse of the Pythagoras theorem.
• Explain why verifying a conjecture for several cases is not the same as proving it, using the region-counting counter-example at n = 6.

Ask the tutor: Can you walk me through writing the negation and converse of one of my own geometry theorems, step by step?