Scientific notation

The distance from Earth to the Sun is about 150,000,000150{,}000{,}000 km. A red blood cell is about 0.0000070.000007 metres wide. Numbers this large or this small are clumsy to write, read, and compare in ordinary decimal form — count the zeros wrong by one and the answer is out by a factor of ten. Scientific notation (also called standard form) fixes this by writing every number as a small decimal between 11 and 1010 multiplied by a power of ten, so 150,000,000150{,}000{,}000 km becomes the tidy 1.5×1081.5\times10^8 km. This article, part of the Class 8 "Power Play" chapter, builds on exponential notation and the laws of exponents already covered, and shows how to convert fluently between standard form and ordinary form, and how standard form makes comparing huge or tiny numbers almost effortless.

The core insight

Any number can be written uniquely as k×10nk\times10^n with 1≤k<101\le k<10, where a positive nn signals a large number and a negative nn signals a small one, and this single convention makes very large and very small numbers compact, comparable, and easy to compute with.

The core idea

Standard form writes a number as a product of two parts: a decimal coefficient with exactly one non-zero digit before the point, and a power of ten that carries all the "size" information.

k×10n,1≤k<10, n∈Zk\times10^n,\qquad 1\le k<10,\ n\in\mathbb{Z}

where: kk is the coefficient (sometimes called the mantissa), restricted to lie between 11 and just under 1010 so the form is unique, and nn is an integer exponent — positive for numbers greater than or equal to 1010, negative for numbers between 00 and 11, and zero only for numbers already between 11 and 1010.

Large numbers get a positive exponent. To convert an ordinary large number, slide the decimal point left until only one non-zero digit remains in front of it, and count how many places it moved — that count is nn.

347,000=3.47×105347{,}000 = 3.47\times10^{5}

The point moved 55 places left (from after the trailing zeros to between 33 and 44), so n=5n=5.

Small numbers get a negative exponent. To convert a decimal smaller than 11, slide the point right until exactly one non-zero digit sits before it; the number of places moved becomes a negative nn.

0.000562=5.62×10−40.000562 = 5.62\times10^{-4}

The point moved 44 places right, so n=−4n=-4.

Converting back to ordinary form reverses the move: a positive exponent pushes the point right (the number grows), a negative exponent pushes it left (the number shrinks), inserting zeros to fill any gaps. The real payoff of standard form is comparison. Two numbers written this way are compared first by their exponents — whichever has the larger nn is the larger number — and only if the exponents tie do the coefficients need comparing. This turns "which is bigger, 0.0004820.000482 or 0.000390.00039?" from a digit-counting exercise into an instant glance at 4.82×10−44.82\times10^{-4} versus 3.9×10−43.9\times10^{-4}. Standard form is why very large or very small measurements — the mass of a planet, the size of a bacterium, the number of stars in a galaxy — are usually quoted this way in science and general knowledge, rather than as long strings of digits: it keeps the precision of a number (how many meaningful digits it has, carried in kk) visibly separate from its size (carried entirely by nn). Two numbers that look wildly different in ordinary form, such as 6,0006{,}000 and 6,400,0006{,}400{,}000, are instantly seen to differ by a factor close to a power of ten once written as 6×1036\times10^3 and 6.4×1066.4\times10^6.

Figure
A number line showing tiny numbers on the left with negative exponents and huge numbers on the right with positive exponents, all expressed in standard form
Standard form spans tiny to huge on one scale 5.62×10⁻⁴ a small decimal 3.47×10⁰ an ordinary number, n = 0 3.47×10⁵ a large number 1.5×10⁸ Earth to Sun, km More negative exponent = smaller number; more positive exponent = larger number
Derivation AConverting a large number to standard form

The goal is to show, step by step and with no guessing, why sliding the decimal point in 52,600,00052{,}600{,}000 leftward until one non-zero digit remains and counting the slides gives the correct exponent nn in k×10nk\times10^n, and to check the result actually reconstructs the original number.

Step 1. Identify the first non-zero digit from the left of 52,600,00052{,}600{,}000: it is 55. Every digit before it (there are none here, since 55 is already the leading digit) would otherwise have to be dropped as the point moves past it.

Step 2. Place the decimal point immediately after this leading digit, keeping the significant digits that follow: 5.265.26. The trailing zeros of the original number are absorbed into the power of ten and dropped from the coefficient.

Step 3. Count how many places the decimal point travelled from its original position at the very end of 5260000052600000 to its new position between the 55 and the 22: it travelled 77 places to the left. Each leftward place is worth one power of ten, since dividing by 1010 each time is exactly what moving the point left by one place represents.

Step 4. Write the result as k×10nk\times10^n with k=5.26k=5.26 and n=7n=7, since the point moved left (large number, positive exponent):

52,600,000=5.26×10752{,}600{,}000 = 5.26\times10^{7}

Step 5. Verify by reversing the move: sliding the point in 5.265.26 seven places to the right regenerates 52,600,00052{,}600{,}000, confirming the conversion is correct and that 5.265.26 indeed satisfies 1≤k<101\le k<10.

Worked example

Write 0.00003810.0000381 in standard form, and convert 9.4×10−39.4\times10^{-3} back to ordinary form.

Step 1. For 0.00003810.0000381, find the first non-zero digit: it is 33. Move the decimal point to sit just after this digit, giving the coefficient 3.813.81.

Step 2. Count the places moved: starting from 0.00003810.0000381, the point travels 55 places to the right to land between the 33 and 88.

Step 3. Since the point moved right (a small number), the exponent is negative:

0.0000381=3.81×10−50.0000381 = 3.81\times10^{-5}

Step 4. Now convert 9.4×10−39.4\times10^{-3} back to ordinary form. The exponent −3-3 means moving the decimal point in 9.49.4 three places to the left, since a negative exponent signals a small number.

Step 5. Moving left one place at a time: 9.4→0.94→0.094→0.00949.4 \to 0.94 \to 0.094 \to 0.0094.

Final answer: 0.0000381=3.81×10−50.0000381 = 3.81\times10^{-5}, and 9.4×10−3=0.00949.4\times10^{-3} = 0.0094.

Competitive stretch

Compare 6.3×10−66.3\times10^{-6} and 9.9×10−79.9\times10^{-7} without converting either to ordinary form. The first has the larger exponent (−6>−7-6 > -7), so it is the larger number — no need to look at the coefficients at all, since exponent comparison alone settles it here. As a further check, notice that 9.9×10−79.9\times10^{-7} is almost 1×10−61\times10^{-6}, so the comparison is really 6.3×10−66.3\times10^{-6} against roughly 1×10−61\times10^{-6} — nowhere close, which confirms the answer without any long-hand conversion to ordinary decimals.

Common traps

  • Letting kk fall outside 1≤k<101\le k<10. Writing 52.6×10652.6\times10^6 instead of 5.26×1075.26\times10^7 is not proper standard form; always adjust until exactly one non-zero digit precedes the decimal point.
  • Getting the sign of the exponent backwards. Numbers greater than 11 need a positive exponent; numbers between 00 and 11 need a negative exponent — mixing these up is the single most common error.
  • Miscounting decimal-point shifts, especially with several zeros in a row — count one place at a time rather than estimating.
  • Forgetting to renormalise kk after a calculation, or comparing coefficients before exponents. If adding exponents produces 12×10512\times10^5, rewrite it as 1.2×1061.2\times10^6; and when judging which of two standard-form numbers is bigger, always check the exponent first — a larger coefficient with a smaller exponent, such as 9.8×1039.8\times10^{3} versus 1.1×1041.1\times10^{4}, can still lose out (1.1×104=11,0001.1\times10^4 = 11{,}000 is bigger than 9,8009{,}800).

Ask the tutor: Stuck on which way to move the decimal point, or want a tricky conversion checked? Ask the AI tutor to work through it with you.

Why does the exponent have to be an integer, not a fraction?

Standard form is built directly from repeatedly multiplying or dividing by 1010, and each such step shifts the decimal point by exactly one whole place. There is no operation in this scheme that corresponds to "half a shift," so the exponent counts a whole number of decimal-point moves and must be an integer. Powers with fractional exponents exist in other areas of mathematics, but they play no role in writing ordinary numbers in standard form.

What happens if the number already starts with a single digit between 1 and 10, like 7.2?

Then no decimal-point movement is needed at all, and the exponent is simply 00, since 100=110^0=1. So 7.27.2 in standard form is 7.2×1007.2\times10^{0}, though in practice it is usually just left as 7.27.2. The rule 1≤k<101\le k<10 is already satisfied, which is exactly why zero is the correct exponent — the number was never "large" or "small" in the first place.

Do I always have to drop the trailing zeros when I write the coefficient?

Yes, for a number like 52,600,00052{,}600{,}000, the zeros after the significant digits 55, 22, and 66 are not written in the coefficient — they are represented entirely by the power of ten, 10710^7. Keeping them, like writing 5.260000×1075.260000\times10^7, would not be wrong arithmetically but is unnecessary; standard form is meant to be a compact way of writing the number, so only the digits that were actually present before the point moved are kept.

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Where this leads