Without Zero You Could Not Tell 105 From 15

To mark an empty place. Without a zero, $105$ and $15$ would look identical — there would be no way to show that the tens place holds nothing.

Zero is what makes place value work without ambiguity. This page covers everything in the CBSE Class 8 Mathematics chapter's second part: converting between bases, place values in any base, the role of zero, and comparing the efficiency of two systems.

Divide repeatedly by the new base, writing down the remainder each time, then read the remainders upward.

Base 10 to base 2. Convert $25$:

- $25÷2=12$, remainder 1
- $12÷2=6$, remainder 0
- $6÷2=3$, remainder 0
- $3÷2=1$, remainder 1
- $1÷2=0$, remainder 1

Reading the remainders from the bottom up: $25=11001_2$.

Base 10 to base 5. Convert $47$:

- $47÷5=9$, remainder 2
- $9÷5=1$, remainder 4
- $1÷5=0$, remainder 1

So $47=142_5$.

Base 10 to base 8. Convert $100$: remainders $4$, $4$, $1$, giving $100=144_8$.

Base 10 to base 7. Convert $100$: remainders $2$, $0$, $2$, giving $100=202_7$.

Base 10 to base 20. Convert $100$: since $100=5\times 20+0$, it is $50_{20}$.

Converting back means multiplying by the place values and adding:

$$11001_2=16+8+0+0+1=25$$
$$142_5=1\times 25+4\times 5+2=25+20+2=47$$
$$144_8=1\times 64+4\times 8+4=64+32+4=100$$

The direction of reading is what students get wrong. The remainders come out least significant first, so they must be read bottom to top — writing them in the order they appeared would give $10011_2$, a different number entirely.

The places are the powers of the base, starting from $n^0=1$ at the right and multiplying by $n$ at each step leftward.

Base 10 — places are $1,10,100,1000$, that is $10^0,10^1,10^2,10^3$.

Base 2 — places are $1,2,4,8,16,32$. So in $11001_2$:

- the leftmost $1$ is in the 16s place, so $1\times 16=16$
- the next $1$ is in the 8s place, so $8$
- the two $0$s contribute nothing
- the last $1$ is in the 1s place, so $1$

$$16+8+1=25$$

Base 5 — places are $1,5,25,125$. In $142_5$, the $1$ is worth $25$, the $4$ is worth $20$ and the $2$ is worth $2$.

Base 8 — places are $1,8,64,512$. In $144_8$, the digits are worth $64$, $32$ and $4$.

Base 20 — places are $1,20,400$.

A rule follows about which digits a base can use: a base-$n$ system uses exactly $n$ digits, from $0$ to $n-1$.

- Base 2 uses only $0$ and $1$
- Base 5 uses $0$ to $4$
- Base 8 uses $0$ to $7$

So $158_5$ is not a valid numeral, because the digits $5$ and $8$ do not exist in base 5. That check is worth running on any answer — if a digit is as large as the base, the conversion has gone wrong.

By giving every empty place a symbol, so no number is ambiguous and every number can be written in exactly one way.

Consider trying to write one hundred and five without a zero. It has 1 hundred, no tens and 5 ones. Writing just the digits that appear gives $15$ — which already means fifteen. There is no way to tell the two apart.

With a zero, the tens place is filled explicitly:

$$105=1\times 100+0\times 10+5$$

and it cannot be confused with $15=1\times 10+5$.

The same problem appears in every positional system, which is why the Mayan system needed its shell symbol for zero.

Zero does two separate jobs, and both should be named:

- As a place holder, marking an empty position so the other digits keep their correct places, as in $105$, $2005$ and $40$
- As a number in its own right, the result of $7-7$, the additive identity since $n+0=n$, and a value that can be added and subtracted

These two roles together are what make the system complete: with ten digits including zero, every whole number however large can be written unambiguously, using no new symbols.

The contrast with a non-positional system explains why zero was not needed earlier. Roman numerals have no zero and do not need one, because an absent value is simply an absent symbol — $CV$ is unambiguous. It is precisely because our system depends on position that an empty position has to be shown.

Write the same quantity in both and count the symbols each needs.

Take the number one hundred:

- Base 2 — $1100100_2$, which needs 7 digits
- Base 5 — $400_5$, which needs 3 digits
- Base 8 — $144_8$, which needs 3 digits
- Base 10 — $100$, which needs 3 digits
- Base 20 — $50_{20}$, which needs 2 digits
- Roman — $C$, which needs 1 symbol for this particular number
- Egyptian — one coil of rope, 1 symbol here
- Tally — 100 strokes

Checking base 2: $64+32+4=100$. Checking base 5: $4\times 25=100$.

The pattern is clear. A larger base needs fewer digits to write a given number, because each place covers a bigger range.

But there is a cost, and this is the real comparison. A larger base needs more distinct symbols to learn — base 20 requires twenty different digits, while base 2 needs only two. So the choice is a trade-off:

- Base 2 — very few symbols, but long numerals
- Base 20 — short numerals, but many symbols to memorise
- Base 10 — a workable middle, with ten symbols and reasonably short numerals

And the Roman and Egyptian entries above are misleading on purpose. They look efficient for $100$ because it happens to be a landmark number — but $248$ needs eight Roman symbols against three digits, and tally marks need a hundred strokes for a number our system writes with three. Efficiency must be judged across many numbers, not one convenient case.