Standard form writes any number as a value between $1$ and $10$, multiplied by a power of $10$ — avoiding long strings of zeros for very large or very small numbers:

Core tier note: calculating with values already in standard form is only examined on Paper 3.

1. Place the decimal point after the first non-zero digit.
2. Count how many places the point moved.
3. If the original number was $10$ or more, the power is positive; if it was less than $1$, the power is negative.

Example (a): $3\,600\,000 \to 3.6 \times 10^6$ (decimal moved $6$ places, number is large so power is positive).

Example (b): $0.00047 \to 4.7 \times 10^{-4}$ (decimal moved $4$ places, number is small so power is negative).

Move the decimal point in the opposite direction: right for a positive power, left for a negative power.

Example: write $9.02 \times 10^{-3}$ as an ordinary number. Move the point $3$ places left:

Multiply (or divide) the front numbers together, and add (or subtract) the powers of $10$ separately. Then re-normalise if the front number falls outside $1 \leqslant A < 10$.

Example: calculate $(5 \times 10^4) \times (6 \times 10^3)$.

1. Multiply the front numbers and add the powers: $(5\times6) \times 10^{4+3} = 30 \times 10^7$.
2. $30$ is not a valid front number, so rewrite it: $30 = 3 \times 10^1$.
3. Combine: $3 \times 10^1 \times 10^7 = 3 \times 10^8$.

Divide the front numbers and subtract the powers of $10$, then re-normalise if needed.

Example: calculate $(8 \times 10^7) \div (2 \times 10^3)$, giving your answer in standard form.

1. Divide the front numbers: $8 \div 2 = 4$.
2. Subtract the powers: $10^{7-3}=10^4$.
3. Combine: $4 \times 10^4$. Since $4$ is already between $1$ and $10$, no re-normalising is needed.

Standard form is especially useful in real contexts where numbers are extreme in size — the mass of a dust particle ($\approx 7.5\times10^{-10}\text{ kg}$) or the distance to a star ($\approx 4\times10^{16}\text{ m}$) would be unmanageable written out in full with every zero.

• Leaving the front number outside the valid range, e.g. stopping at $30 \times 10^7$ without re-normalising to $3 \times 10^8$.
• Getting the sign of the power wrong — small numbers need a negative power, large numbers need a positive one.
• Losing count of the decimal shifts when there are many zeros — count one digit at a time rather than guessing from the zeros.