The World of Numbers

This is a direct proportion, so first find the ingots for a single bag.

12 bags is 6 groups of 2 bags, and each group earns 15 ingots.

The merchant receives 90 copper ingots.

Test each number for factors other than 1 and itself:

Every one of them is a prime number. More precisely, they are all the prime numbers lying between 10 and 20.

Next three in the pattern. Continuing with the primes after 19:

(21 = 3×7, 25 = 5×5 and 27 = 3×9 are skipped because they are composite.)

A set is closed under an operation if performing that operation on any two members of the set always gives a result that is still inside the set.

Addition. For any two natural numbers the sum is again a natural number:

So \(\mathbb{N}\) is closed under addition.

Subtraction. Sometimes the answer stays inside \(\mathbb{N}\):

But reversing the same two numbers does not:

−5 is not a natural number. A single counter-example is enough, so natural numbers are not closed under subtraction.

This exact gap is what forced mathematics to invent the negative numbers, and it is the story taken up in the next section of this chapter.

The thumb does the counting, so it is not itself counted. That leaves 4 fingers, each with 3 joints:

So 12 can be counted on a single hand.

Link to base 12. A counting system naturally takes as its base the largest number one full cycle of the hand can hold. One sweep of the thumb across the finger joints ends at 12, so the count ‘rolls over’ at 12 — exactly what a base-12 (duodecimal) system does.

And base 60. The other hand can record how many complete twelves have gone by. With 5 digits on that hand:

which is the origin of the Babylonian base-60 (sexagesimal) system — and the reason an hour still has 60 minutes and a circle 360 degrees.

A drop is a subtraction:

The temperature at midnight was −11 °C, that is, 11 °C below zero.

On the number line you start at 4 and move 15 units to the left, crossing zero and landing 11 units past it.

Writing each event as an integer. A debt or loss is negative, a profit is positive:

The equation is:

Adding step by step:

His final standing is −₹100 — he still owes ₹100. Brahmagupta would have called this a ṛṇa (debt) rather than a dhana (fortune).

(i) A fortune multiplied by a debt is a debt, so the answer is negative:

(ii) A debt multiplied by a debt is a fortune, so the answer is positive:

(iii) Subtracting a debt from zero leaves a fortune:

(iv) A debt divided by a fortune is a debt:

The situation. Suppose you have ₹10 in your pocket and, written in your notebook, a debt of ₹5 that you owe a friend. Your true worth is:

Now remove the debt. Your friend forgives the ₹5. Removing a debt of 5 is written as \(-(-5)\), so your worth becomes:

But cancelling a ₹5 debt leaves you exactly ₹5 better off — the same as if someone had handed you ₹5:

Why it must be so. Taking away something negative removes a burden, and removing a burden is a gain. This is why the two minus signs combine into a plus.

On the number line, subtracting means facing left; subtracting a negative reverses that direction, so you end up moving right.

Two rational numbers \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) are equal exactly when \(a\times d=b\times c\) (cross multiplication).

(i)

The cross products agree, so \(\dfrac{2}{3}=\dfrac{4}{6}\).

(ii)

So \(\dfrac{5}{4}=\dfrac{10}{8}\).

(iii)

So \(-\dfrac{3}{5}=-\dfrac{6}{10}\).

(iv) Write 3 as \(\dfrac{3}{1}\):

So \(\dfrac{9}{3}=3\).

To add rational numbers, rewrite them with the LCM of the denominators.

(i) LCM of 5 and 10 is 10:

(ii) LCM of 12 and 8 is 24:

(iii) LCM of 7 and 14 is 14:

Answers: (i) \(\dfrac{7}{10}\)  ·  (ii) \(\dfrac{29}{24}\)  ·  (iii) \(-\dfrac{5}{14}\)

(i) LCM of 6 and 4 is 12:

(ii) LCM of 8 and 4 is 8:

(iii) Subtracting a negative is adding a positive:

Answers: (i) \(\dfrac{7}{12}\)  ·  (ii) \(\dfrac{5}{8}\)  ·  (iii) \(-\dfrac{1}{9}\)

For a product, multiply the numerators together and the denominators together, then reduce.

(i)

(ii) 7, 11, 5 and 8 share no common factor, so nothing cancels:

(iii) A negative times a positive is negative:

Answers: (i) \(\dfrac{1}{5}\)  ·  (ii) \(\dfrac{35}{88}\)  ·  (iii) \(-\dfrac{10}{49}\)

To divide by a rational number, multiply by its reciprocal.

(i)

(ii)

(iii)

Answers: (i) \(\dfrac{20}{9}\)  ·  (ii) \(\dfrac{56}{55}\)  ·  (iii) \(-\dfrac{8}{5}\)

This is the distributive property of multiplication over addition. Evaluate each side separately.

Left-hand side.

Right-hand side.

Both sides equal \(\dfrac{10}{3}\), so the statement is proved.

Method 1 — bracket first.

Method 2 — distribute first. Multiplication distributes over subtraction as well:

Both routes give \(\dfrac{1}{12}\), which is what the distributive property guarantees.

Expand the left-hand side using the distributive property:

Work out that last product:

So the left-hand side is:

which is exactly the right-hand side. Subtracting one side from the other leaves:

Conclusion. The two sides are identical for every value of \(x\). This is not an equation with one solution — it is an identity, and every rational number \(x\) satisfies it.

First write every number in the same form so their positions can be compared:

Marking each one.

Order on the line, from left to right:

Method. Rewrite both numbers with a common denominator, then pick numerators in between.

The LCM of 2 and 4 is 4:

Only \(0\) and \(-\frac{1}{4}\) sit between these as quarters, so enlarge the denominator to 8:

Now any numerator from −3 to 1 works. Choosing three of them:

Check. In decimals the bounds are \(-0.5\) and \(0.25\), and:

All three lie strictly between the given numbers, so \(-\dfrac{3}{8},\ 0,\ \dfrac{1}{8}\) is a valid answer. Many other answers are equally correct.

The LCM of 4 and 12 is 12:

Reduce by dividing both parts by 2:

The answer is \(\dfrac{1}{6}\).

Convert both mixed numbers to improper fractions:

The number of kurtas is the total length divided by the length of one kurta:

He can make exactly 7 kurtas.

Method. Two decimals that differ in the 4th place have room for infinitely many numbers in the 5th place. Write both with five decimal places:

Any five-place decimal with digits strictly between these works. Three of them are:

As fractions, these are:

so all three are genuinely rational numbers.

Yes — here are three reliable methods.

Method 1: take the average (the mean). For any two rationals \(a\) and \(b\) with \(a<b\):

The average of two rationals is always rational, so this always works. For \(a=\frac{1}{3}\) and \(b=\frac{1}{2}\):

Method 2: enlarge the common denominator. Write both as fractions over a common denominator, then multiply top and bottom of both by 10 (or 100). This opens up gaps in the numerators:

Now \(\frac{21}{60},\frac{22}{60},\dots,\frac{29}{60}\) are all strictly in between — nine of them at once.

Method 3: compare the decimals. Convert both to decimals and change a digit far enough to the right. Between 0.333… and 0.5, the number 0.4 is obviously in between.

The test. Write the fraction in lowest terms and factorise the denominator. The decimal terminates if and only if the denominator's only prime factors are 2 and 5 — because only then can it be rewritten with a denominator that is a power of 10.

(i) \(\dfrac{7}{20}\):

Only 2s and 5s, so it terminates. Checking:

(ii) \(\dfrac{4}{15}\):

The factor 3 is neither 2 nor 5, so it repeats. Checking:

(iii) \(\dfrac{13}{250}\):

Only 2s and 5s, so it terminates. Checking:

The long division. Dividing 1 by 13 gives:

The repeating block is 076923, six digits long. (It cannot be longer than 12, since only the remainders 1 to 12 are possible.)

Now the multiples.

What we notice. There are two cyclic families, not one:

Family A holds \(k=1,3,4,9,10,12\) and family B holds \(k=2,5,6,7,8,11\). Within each family the digits are just rotations of one another — so 13 does show cyclic behaviour, but in two separate cycles.

Contrast with \(\dfrac{1}{7}\), whose block 142857 has length 6 = 7 − 1. There all six multiples are rotations of the same block, so \(\frac{1}{7}\) is a true cyclic number. For 13 the block length is 6, only half of 12, which is exactly why the multiples split into two families.

(i) \(\sqrt{81}\) is a perfect square:

Rational.

(ii) 12 is not a perfect square:

Since \(\sqrt{3}\) is irrational, so is \(2\sqrt{3}\). Irrational.

(iii) A repeating decimal. Let \(x=0.333\ldots\), then \(10x=3.333\ldots\) and:

Rational, equal to \(\dfrac{1}{3}\).

(iv) The block 12345 repeats, so let \(x=0.\overline{12345}\) and multiply by \(10^{5}\):

Rational.

(v) Here the number of zeros grows by one each time — 1.01, then 001, then 0001, and so on. No fixed block ever repeats, so the expansion is non-terminating and non-repeating. Irrational.

(vi) This decimal stops after 21 places, and every terminating decimal is rational:

Rational. (It can be reduced, but any correct \(\frac{p}{q}\) form is enough.)

Let:

Multiply both sides by 10. The digits after the point are unchanged, because the 9s go on forever:

Now subtract the first equation from the second. The infinite tails of 9s cancel exactly:

So \(0.\overline{9}=1\) — not ‘almost 1’, but exactly 1.

A second way to see it.

and the left-hand side is plainly 1.

What to look for. The block of \(\frac{1}{n}\) is fully cyclic when its length is exactly \(n-1\), the longest it could possibly be. Such primes are called full reptend primes.

Testing small primes:

So the next such numbers after 7 are:

An example. For \(n=17\) the block is 0588235294117647, and multiplying it by 2, 3, …, 16 simply rotates those same sixteen digits — exactly the behaviour of 142857.

(i) \(\dfrac{3}{50}\). The denominator is \(50=2\times 5^{2}\), only 2s and 5s, so expect a terminating decimal:

Terminating, two decimal places.

(ii) \(\dfrac{2}{9}\). The denominator 9 = 3² contains a 3, so expect a repeating decimal. Long division gives a remainder of 2 at every step:

Non-terminating and repeating, with a one-digit block.

We use proof by contradiction — the same method Hippasus used for \(\sqrt{2}\).

Step 1: assume the opposite. Suppose \(\sqrt{5}\) is rational. Then it can be written as

where \(p\) and \(q\) are integers, \(q\neq 0\), and the fraction is in lowest terms — so \(p\) and \(q\) share no common factor.

Step 2: square both sides.

Step 3: deduce that 5 divides \(p\). The right-hand side is a multiple of 5, so \(p^{2}\) is a multiple of 5. Since 5 is prime, this forces \(p\) itself to be a multiple of 5. Write:

Step 4: substitute back.

Step 5: deduce that 5 divides \(q\) too. By the same argument, \(q^{2}\) is a multiple of 5, so \(q\) is a multiple of 5.

Step 6: the contradiction. Both \(p\) and \(q\) are multiples of 5. But we assumed the fraction was in lowest terms, so they had no common factor. This is impossible.

The only assumption we made was that \(\sqrt{5}\) is rational, so that assumption must be false. Hence \(\sqrt{5}\) is irrational. \(\blacksquare\)

The general method. Let \(x\) be the number. Multiply by \(10^{m}\) where \(m\) is the count of non-repeating decimal digits, then by \(10^{n}\) more where \(n\) is the length of the repeating block, and subtract the two results. The infinite tails cancel.

(i) Terminating, one decimal place:

(ii) Terminating, four decimal places:

(iii) \(3.0525252\ldots\) — one non-repeating digit, block length 2. Let \(x=3.0\overline{52}\):

(iv) \(1.2353535\ldots\) — one non-repeating digit, block length 2. Let \(x=1.2\overline{35}\):

(v) \(0.232323\ldots\) — no non-repeating digits, block length 2. Let \(x=0.\overline{23}\):

(vi) \(2.05555\ldots\) — one non-repeating digit, block length 1. Let \(x=2.0\overline{5}\):

(vii) \(2.125555\ldots\) — two non-repeating digits, block length 1. Let \(x=2.12\overline{5}\):

(viii) \(3.125555\ldots\) — identical working, only the whole part differs:

(ix) \(2.162516251625\ldots\) — no non-repeating digits, block length 4. Let \(x=2.\overline{1625}\):

Both are located by successive magnification — zoom into a smaller and smaller interval at each step.

(i) \(0.532\)

Three magnifications are enough, because the decimal has three places and then stops.

(ii) \(1.1\overline{5}=1.15555\ldots\) First convert it to a fraction. Let \(x=1.1\overline{5}\):

Now magnify:

The 5s never stop, so the zooming never finishes — but each step traps the number in an interval ten times smaller, closing in on the single point \(\dfrac{52}{45}\).

Method. To fit 6 numbers between two integers, write both with a denominator of \(6+1=7\). That leaves exactly six gaps.

The numerators strictly between 21 and 28 are 22, 23, 24, 25, 26 and 27, giving:

Check. In decimals these are approximately 3.14, 3.29, 3.43, 3.57, 3.71 and 3.86 — all strictly between 3 and 4.

Simpler answers such as 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 are equally correct. There are infinitely many valid answers.

The denominators already match, but the numerators 2 and 3 are consecutive — no room in between. So enlarge the denominator by multiplying top and bottom of both by 6:

Now the numerators 13, 14, 15, 16 and 17 all lie strictly in between:

In lowest terms:

Check. As decimals: 0.4 < 0.433 < 0.467 < 0.5 < 0.533 < 0.567 < 0.6. \(\checkmark\)

Step 1: a common denominator. The LCM of 6 and 5 is 30:

Step 2: read off the numerators in between. The integers strictly between 5 and 12 are 6, 7, 8, 9, 10 and 11 — six of them, and we need five:

In lowest terms:

Check. As decimals the bounds are 0.1667 and 0.4, and the five numbers are 0.2, 0.233, 0.267, 0.3 and 0.333 — all strictly inside. \(\checkmark\)

Combine the left-hand side over the LCM of 3 and 5, which is 15:

So the equation becomes:

The denominators are equal, so the numerators must be:

Check. Substituting \(x=2\):

Start from the given condition and isolate \(a\):

Now multiply both sides by \(b\) (allowed, since \(b\neq 0\)):

So \(ab\) is not merely negative — it is exactly −1, whatever non-zero values \(a\) and \(b\) take.

Justification in words. The condition says \(a\) is the negative of the reciprocal of \(b\). A number times its own reciprocal is 1, so a number times the negative of its reciprocal is −1. Since −1 < 0, \(ab\) is negative.

Part 1: the form \(\dfrac{p}{10^{4}}\).

The expansion stops at the 4th decimal place, so the number looks like \(0.d_1d_2d_3d_4\) (possibly with a whole part). Multiplying by \(10^{4}\) shifts the point four places right and leaves an integer. Call it \(p\):

The last non-zero digit is \(d_4\), so \(d_4\neq 0\). But \(d_4\) is the units digit of \(p\), so \(p\) does not end in 0 — that is, \(p\) is not divisible by 10. \(\blacksquare\)

Part 2: the lowest-form denominator. Yes — it is always divisible by \(2^{4}\) or by \(5^{4}\) (at least one of the two). Here is why.

Since \(p\) is not divisible by 10, it cannot have both 2 and 5 as factors. Two cases arise.

Case 1: \(p\) is odd. Then \(p\) shares no factor of 2 with \(10^{4}=2^{4}\times 5^{4}\). Only 5s can cancel, so the full \(2^{4}\) survives:

which is divisible by \(2^{4}\). Example: \(0.1235=\dfrac{1235}{10^{4}}=\dfrac{247}{2000}\), and \(2000=2^{4}\times 5^{3}\).

Case 2: \(p\) is even. Then \(p\) is not divisible by 5, so only 2s can cancel and the full \(5^{4}\) survives:

which is divisible by \(5^{4}\). Example: \(0.0002=\dfrac{2}{10^{4}}=\dfrac{1}{5000}\), and \(5000=2^{3}\times 5^{4}\).

In both cases at least one of \(2^{4}\) and \(5^{4}\) divides the denominator, so the answer is yes, it is necessary.

Step 1: check the fraction is in lowest terms. \(18=2\times 3^{2}\) and \(125=5^{3}\) share no common factor, so \(\dfrac{18}{125}\) is already reduced.

Step 2: factorise the denominator.

The only prime factor is 5, so the expansion terminates.

Step 3: count the decimal places. Write the denominator as \(2^{m}\times 5^{n}\); here \(m=0\) and \(n=3\). The number of decimal places is \(\max(m,n)=3\).

Confirming without long division — multiply top and bottom by \(2^{3}=8\) to make the denominator a power of 10:

Terminating, 3 decimal places.

The rule. If the lowest-form denominator is \(2^{m}\times 5^{n}\), the decimal terminates after exactly \(\max(m,n)\) places.

Here \(m=3\) and \(n=1\):

So the expansion has 3 decimal places.

Why. To turn the denominator into a power of 10 we need equal counts of 2s and 5s. There are three 2s but only one 5, so we supply two more 5s:

The denominator becomes \(10^{3}\), giving three decimal places. Example:

Step 1: the obvious common denominator is not enough. With \(m=12\):

Here \(k_2-k_1=10-7=3\), which is not greater than 6. Only two integers (8 and 9) sit strictly in between — too few for five numbers.

Step 2: enlarge \(m\). Multiply top and bottom of both by 5, giving \(m=60\):

Step 3: pick five numerators strictly between 35 and 50. Taking 36, 38, 40, 42 and 44:

In lowest terms:

Check. As decimals: \(a=0.583\), then 0.6, 0.633, 0.667, 0.7, 0.733, then \(b=0.833\). All five are strictly between. \(\checkmark\)

Step 4: why the condition is needed. The method only produces numbers of the form \(\dfrac{k}{m}\) with \(k\) an integer strictly between \(k_1\) and \(k_2\). The count of such integers is:

To obtain \(n\) of them we therefore need:

i.e. \(k_2-k_1\ge n+1\). Requiring the strict inequality \(k_2-k_1>n+1\) leaves a spare gap and guarantees at least \(n\) choices with room to spare. For \(n=5\) this gives \(k_2-k_1>6\), which is exactly the condition stated in the question.

The key identity. For any three numbers:

Rearranging it to make the sum of squares the subject:

Substitute the two given conditions. Both brackets are zero:

Finish the argument. The square of any rational number is either positive or zero — never negative. So \(x^{2}\ge 0\), \(y^{2}\ge 0\) and \(z^{2}\ge 0\).

If even one of them were non-zero, its square would be strictly positive, and a sum of non-negative terms with at least one positive term cannot be zero. So every term must vanish:

and therefore:

Take \(a\neq b\), and without loss of generality suppose \(a<b\). (If \(b<a\), swap the names.)

First, it is bigger than \(a\). Add \(a\) to both sides of \(a<b\):

Divide by 2, which is positive so the inequality direction is unchanged:

Second, it is smaller than \(b\). Add \(b\) to both sides of \(a<b\):

Dividing by 2:

Combining the two results:

It is also rational. The sum of two rationals is rational, and a rational divided by 2 is rational. So \(\dfrac{a+b}{2}\) is a rational number lying strictly between \(a\) and \(b\).

How the spiral is built. The first triangle has two perpendicular sides of length 1. Each new triangle then uses the previous hypotenuse as one leg and a fresh segment of length 1 as the other.

Triangle 1. Legs 1 and 1, so by the Pythagoras theorem:

Triangle 2. Legs \(\sqrt{2}\) and 1:

Triangle 3. Legs \(\sqrt{3}\) and 1:

The general rule. Each step adds exactly 1 under the root sign, so the \(n\)-th hypotenuse is:

For the 10 triangles shown in Fig. 3.14:

So the hypotenuses are \(\sqrt{2},\sqrt{3},\sqrt{4},\ldots,\sqrt{11}\).