Class 9 Mathematics Chapter 9: Propositions and their Converses

A proposition is a sentence you can call true or false. 'Draw the altitude from the third vertex' is not a proposition, because it is an instruction and cannot be true or false.

Converse: if the two diagonals of a quadrilateral are equal in length, then it is a square. This is false — a rectangle that is not a square has diagonals of equal length.

It must fit the if-part and fail the then-part. For 'if the queue at the stop grows, then the bus is late', a school finishing early gives a long queue with the bus on time, so it fits the if and fails the then.

'If X then Y' claims Y follows every time X holds, so a single case where it does not contradicts the claim completely. Agreeing cases only cover the cases tested, and the next one might disagree.

Fermat claimed every number of the form $2^{2^n} + 1$ is prime. It holds for $n = 0, 1, 2, 3$, giving 3, 5, 17 and 257, but Euler showed that at $n = 5$ the number $2^{32} + 1$ is composite.

If a number is a multiple of 10, then it is a multiple of 5 — true, because $10 = 2 \times 5$ so the factor 5 is carried. The converse, if a number is a multiple of 5 then it is a multiple of 10, is false: 15 is a multiple of 5 and not of 10.

The squares converse is false: $x = 2$, $y = -2$ gives $x^2 = y^2$ without $x = y$. The cubes converse is true, because every real number has exactly one real cube root, so $x^3 = y^3$ forces $x = y$. Squaring loses the sign and cubing does not.

The original is true: a multiple of 36 inherits the factors 6 and 12. The converse, if $n$ is divisible by both 6 and 12 then it is divisible by 36, is false — $n = 12$ is divisible by 6 and by 12 but not by 36.

'Y when X' means the same as 'if X then Y'. Here X is 'the number ends in the digit 0' and Y is 'it is divisible by 5', so ending in the digit 0 is what you are given.

The converse swaps the two halves and gives 'if $Y$ then $X$'. The statement 'if not $X$ then not $Y$' negates both halves instead of swapping them, so it is a different statement.

The converse of 'if P then Q' is 'if Q then P', formed by swapping the two halves and negating nothing; a sentence that only sets the scene stays where it is. Whether the converse is true has to be settled separately from the original.

First example: if a triangle has two sides of the same length, then the angles facing those sides match. Its converse — if a triangle has two angles of the same size, then the sides facing them match — is also true, so here both directions hold.

Second example: if a number is a multiple of 10, then it is a multiple of 5. This is true, since $6 = 2 \times 3$. Its converse — if a number is a multiple of 5, then it is a multiple of 10 — is false, because 9 is a multiple of 5 and not of 6.

So knowing a proposition is true tells you nothing by itself about its converse.

A proposition of the form 'if X then Y' claims that Y follows every time X holds. A counterexample is a single case that fits the if-part and fails the then-part, and because the claim was universal that one case refutes it completely. The reverse does not hold: any number of agreeing cases leaves the untested cases open.

Fermat claimed that every number of the form $2^{2^n} + 1$ is prime. At $n = 0, 1, 2, 3$ the numbers are 3, 5, 17 and 257, and all four are prime. Euler then took $n = 5$ and showed that $2^{32} + 1$ is composite. Four agreeing cases had not made the claim true, and one disagreeing case ended it.

This is why finding a counterexample counts as real mathematical work: it settles a question in a line where a proof might take a page.

The opening sentence sets the scene and is neither half, so it stays exactly as it is. Swapping the if-part with the then-part inside the second sentence gives: suppose a triangle has sides of length $a$, $b$ and $c$. If $a^2 + b^2 = c^2$, then the triangle is right-angled. Deciding it is a separate question from the original. One cannot assume it from the theorem, because a true proposition may have a false converse. What has to be shown is that every triangle whose sides satisfy $a^2 + b^2 = c^2$ must in fact be right-angled. The route is to construct a right triangle with perpendicular sides of lengths $a$ and $b$, use the theorem on it to find its third side, and then compare the two triangles. All three pairs of sides match, so by the SSS criterion they are congruent, and the original triangle must have been right-angled too. The converse is therefore true.

There are two separate mistakes.

The first is treating agreeing examples as a proof. 'If X then Y' claims Y follows in every case where X holds, so eight agreeing cases establish nothing about the cases not tested. Fermat's claim held for four values of $n$ and still failed at $n = 5$, where Euler showed $2^{32} + 1$ is composite. Only a proof settles a proposition; examples can refute one but never establish it.

The second is assuming the converse follows. The converse is a different statement, obtained by swapping the halves, and its truth has to be settled on its own. 'If a number is a multiple of 10, then it is a multiple of 5' is true, while its converse is false because 9 is a multiple of 5 and not of 6.

The correct procedure is to try to break each statement with a counterexample, and to look for a proof of whichever one survives.

It must begin at X, which is what you are allowed to assume, and end at Y, which is what is claimed. Beginning at Y and ending at X would prove the converse instead.

The companion of a divisor is the number divided by that divisor, which is itself a divisor. For 12 the couples are 1 with 12, 2 with 6, and 3 with 4.

If every divisor had a different companion, the divisors would fall into couples of two and the count would be even. Since the count is odd, at least one divisor must be its own companion, so the number is that divisor multiplied by itself, which makes it a square.

It proves that an odd count of divisors leads to the number being a square, because that is the direction it travels. It leaves open whether every square has an odd count of divisors, which is the converse and needs a separate argument.

In a square, one divisor is its own companion, and there is only one such divisor. Every other divisor pairs with a different one, so those come in couples and contribute an even count. Adding the one that stands alone makes the total odd.

It means each condition brings the other, so they describe exactly the same cases. It is written with the phrase ‘if and only if'.

A whole number is even if and only if its last digit is even: both directions hold, so the conditions are equivalent. For both failing, take ‘if a number is prime then it is odd', which fails at 2, and its converse ‘if a number is odd then it is prime', which fails at 9.

The statement is true: congruent triangles match exactly, so every measurement including the area agrees. The converse is false. A triangle with base 6 and height 2 and one with base 3 and height 4 both have area 6, and they are not congruent.

The claim covers every case, and the eight say nothing about the cases that were not checked. A single untested case could still fail.

The chain shows only that each step leads to the next in that order. ‘B leads to Y' carries no information about whether Y leads back to B. To get the converse every arrow would have to be checked in reverse, and any one of them may fail.

Every divisor of a number comes with a companion: divide the number by the divisor and the result is a divisor as well. For 12 the couples are 1 with 12, 2 with 6, and 3 with 4.

First direction. Suppose the count of divisors is odd. If every divisor had a different companion, the divisors would fall into couples of two and the count would be even. So at least one divisor is its own companion, which means the number is that divisor multiplied by itself, and the number is a square. This argument begins at the odd count, so what it proves is that an odd count of divisors leads to a square number.

It has not proved the converse. Reading the steps backwards is not permitted, because each step was only ever claimed in one direction.

Second direction. Suppose the number is a square. Then one divisor is its own companion, and there is exactly one such divisor, since two different self-paired divisors would give the number two different values. Every other divisor pairs with a different one, so those contribute an even count, and the one standing alone makes the total odd.

Both directions hold, so being a square and having an odd count of divisors are equivalent conditions.

An argument proves ‘if X then Y' only when it begins by assuming X and finishes at Y. If it begins at Y and finishes at X it has proved ‘if Y then X', the converse, which is a different claim about different cases.

The mistake is easy because the two statements are built from the same two halves and look almost identical on the page. It is also easy because one direction is often much simpler than the other, so a student who sets out to prove the hard direction can drift into proving the easy one without noticing.

The check is mechanical: read the first line of the argument. Whatever is assumed there is the if-part of whatever has actually been proved. If that is not the if-part of the statement you were asked about, the argument has settled the converse.

Two conditions are equivalent when each brings the other, so they pick out exactly the same cases and either description can replace the other. Both directions have to be established, which means two arguments, and the phrase ‘if and only if' records that.

An equivalent pair: a whole number is even if and only if its last digit is even. An even number must end in one of 0, 2, 4, 6 or 8, and any number ending in one of those is even, so each condition brings the other.

A pair that is not equivalent: being a square and having an even count of divisors. A square has an odd count, so the two conditions never hold together at all, and neither direction leads to the other.

Both true. A whole number is even exactly when its last digit is even. Any even number ends in 0, 2, 4, 6 or 8, and any number with such a last digit is even, so each direction holds and the conditions are equivalent.

One true only. ‘If two triangles are congruent then their areas are equal' is true, because congruent triangles agree in every measurement. Its converse fails: a triangle with base 6 and height 2 and one with base 3 and height 4 both have area 6 without being congruent.

Both false. ‘If a number is prime then it is odd' fails at 2, which is prime and even. Its converse ‘if a number is odd then it is prime' fails at 9, which is odd and not prime. Each direction needed its own counterexample.

So the truth of a proposition tells you nothing at all about the truth of its converse; it has to be settled separately every time.

Take a triangle whose sides have lengths $a$, $b$ and $c$, with $c$ the side facing the right angle. If the triangle is right-angled, then $a^2 + b^2 = c^2$.

Take a triangle whose sides have lengths $a$, $b$ and $c$. If $a^2 + b^2 = c^2$, then the triangle is right-angled, with $c$ facing the right angle. The scene-setting sentence stays where it is.

Because it is neither the if-part nor the then-part. It only says what $a$, $b$ and $c$ refer to, so swapping it would leave a sentence with no meaning.

The theorem begins at the right angle and ends at $a^2 + b^2 = c^2$, so that is the only direction it settles. The converse begins at $a^2 + b^2 = c^2$ and ends at the right angle, which is a different journey and needs its own argument.

It is applied to the constructed triangle, which was built with a right angle, so the theorem is available for it. It cannot be applied to the given triangle, because whether that triangle has a right angle is precisely what is being established.

Square the longest side on its own: $17^2 = 289$. Add the other two squares: $64 + 225 = 289$. The two agree, so by the converse the triangle is right-angled.

The longest side gives $36$, while the other two squares add to $16 + 25 = 41$. Since 41 is not 36, the triangle is not right-angled.

Three whole numbers satisfying $a^2 + b^2 = c^2$. For example 5, 12, 13 because $25 + 144 = 169$, and 8, 15, 17 because $64 + 225 = 289$.

Suppose $a^2 + b^2 = c^2$ and multiply each side length by $k$. Then $(ka)^2 + (kb)^2 = k^2a^2 + k^2b^2 = k^2(a^2 + b^2) = k^2c^2 = (kc)^2$. So the scaled lengths pass the test as well. Doubling 3, 4, 5 gives 6, 8, 10, and $36 + 64 = 100$.

Because both directions hold. The theorem says a right angle produces the relation, and the converse says the relation produces a right angle. The two conditions are therefore equivalent, so the relation can be used to decide the angle either way.

Let triangle $ABC$ have sides of length $a$, $b$ and $c$ with $a^2 + b^2 = c^2$. Nothing is known about its angles yet — that is what we have to establish.

Construct a second triangle $PQR$ deliberately: place a right angle at $Q$ and make the two sides meeting there have lengths $a$ and $b$. This triangle is right-angled because it was built that way, so the theorem may be applied to it.

By the theorem, its third side satisfies $PR^2 = a^2 + b^2$. We were given $a^2 + b^2 = c^2$, so $PR^2 = c^2$ and therefore $PR = c$, taking the positive length in each case.

Now compare the triangles. Their sides match in three pairs: $a$ with $a$, $b$ with $b$, and $c$ with $c$. By the SSS criterion the two triangles are congruent.

Congruent triangles agree in every angle. $PQR$ has a right angle at $Q$, so $ABC$ has a right angle in the matching position. Hence $ABC$ is right-angled, which is what the converse claims.

The essential move is that the theorem was applied to the triangle we constructed, not to the one we were handed, whose angles were unknown throughout.

A statement sometimes opens with a sentence whose only job is to introduce the letters. In the theorem that sentence is ‘take a triangle whose sides have lengths $a$, $b$ and $c$'. It makes no claim, so it is neither the if-part nor the then-part.

To form the converse, leave that sentence untouched and swap the two halves of the statement that follows. The theorem's second sentence is ‘if the triangle is right-angled, then $a^2 + b^2 = c^2$', so the converse reads: take a triangle whose sides have lengths $a$, $b$ and $c$; if $a^2 + b^2 = c^2$, then the triangle is right-angled.

If the opening sentence is swapped as well, the result is nonsense. The letters would be used in the first half before anything had said what they refer to, and there would be no triangle introduced at all. The test for which sentences may move is simple: only a sentence that asserts something can be an if-part or a then-part.

For 3, 4, 5 the two shorter squares add to $9 + 16 = 25$, and the longest side gives $5^2 = 25$. The two agree, so the test is passed.

For 5, 12, 13 the shorter squares add to $25 + 144 = 169$, and $13^2 = 169$. Again they agree.

To generate more, multiply all three numbers by the same whole number $k$. If $a^2 + b^2 = c^2$ then $(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2c^2 = (kc)^2$, so the scaled triple passes too. Doubling 3, 4, 5 gives 6, 8, 10, where $36 + 64 = 100$; tripling it gives 9, 12, 15, where $81 + 144 = 225$. Doubling 5, 12, 13 gives 10, 24, 26, where $100 + 576 = 676$.

These scaled triples describe triangles of the same shape at a larger size, which is why the right angle survives.

For the first, the longest side is 15, so square it on its own: $15^2 = 225$. The other two squares add to $81 + 144 = 225$. The two numbers agree, so by the CONVERSE of the theorem the triangle is right-angled. Notice that 9, 12, 15 is 3, 4, 5 multiplied by 3.

For the second, the longest side is 6, giving $36$, while the other two squares add to $16 + 25 = 41$. Since 41 and 36 are different, the triangle is not right-angled.

The converse is the part being used in both cases, because each question hands over the side lengths and asks about the angle. The theorem itself runs the other way and would be the part to use if the right angle were given and a side length were wanted.

In each case the longest side is the one squared alone, since that is the side that would face the right angle.

A proposition is a sentence you can call true or false. 'Draw the altitude from the third vertex' is not a proposition, because it is an instruction and cannot be true or false.

Converse: if the two diagonals of a quadrilateral are equal in length, then it is a square. This is false — a rectangle that is not a square has diagonals of equal length.

It must fit the if-part and fail the then-part. For 'if the queue at the stop grows, then the bus is late', a school finishing early gives a long queue with the bus on time, so it fits the if and fails the then.

It must begin at X, which is what you are allowed to assume, and end at Y, which is what is claimed. Beginning at Y and ending at X would prove the converse instead.

The companion of a divisor is the number divided by that divisor, which is itself a divisor. For 12 the couples are 1 with 12, 2 with 6, and 3 with 4.

If every divisor had a different companion, the divisors would fall into couples of two and the count would be even. Since the count is odd, at least one divisor must be its own companion, so the number is that divisor multiplied by itself, which makes it a square.

Take a triangle whose sides have lengths $a$, $b$ and $c$, with $c$ the side facing the right angle. If the triangle is right-angled, then $a^2 + b^2 = c^2$.

Take a triangle whose sides have lengths $a$, $b$ and $c$. If $a^2 + b^2 = c^2$, then the triangle is right-angled, with $c$ facing the right angle. The scene-setting sentence stays where it is.

Because it is neither the if-part nor the then-part. It only says what $a$, $b$ and $c$ refer to, so swapping it would leave a sentence with no meaning.