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If I take a digit like $5$ (say in m) and multiply it by a digit say $5$ (in m) then I will get $25$ ($\mathrm{m}^2$---an area). Now, $5$ can be expressed in Scientific notation as $5 \times 10^0$ to show that it has $1$ significant digit. But then what would the product result $25$ be represented as when expressed to $1$ significant figure using the Scientific notation? $3 \times 10^1$ means the product answer is $30$ and is not $25$. Now $30$ m is far from the correct $25 \; \mathrm{m}^2$.

I have seen the rule that a product of two numbers cannot have more significant digits than any of the multiplicand or the multiplier. And I have read that it is good to use the Scientific notation when working with significant figures to avoid spurious significant digits.

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  • $\begingroup$ Keep in mind that the use of significant figures only applies to measurements. If you want to determine the area to two significant figures, you should measure each length to at least two significant figures. $\endgroup$ – N. F. Taussig Mar 24 '15 at 13:01
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$30\ \mathrm{m}^2$ is correct to one significant figure. The fact that it's far from $25$ is simply because one significant figure isn't very precise.

Scientific notation isn't particularly helpful with numbers such as $25$ or $30$. It becomes more useful when we want to express very small or large numbers succinctly. As an example, $3\times 10^{21}$ is easier to read than $3000000000000000000000$.

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  • $\begingroup$ It looks like you meant $30~\text{m}^2$. $\endgroup$ – N. F. Taussig Mar 24 '15 at 12:53
  • $\begingroup$ Good point, fixed it. $\endgroup$ – Esteemator Mar 24 '15 at 12:54
  • $\begingroup$ Thank you. Yes indeed Scientific notation may not be useful in my question but I used it as it helps to express concisely the number of significant digits that we want. $\endgroup$ – user11206 Mar 24 '15 at 17:19
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Suppose for the moment that measuring $5$ m could somehow contain a possible error whereby a more precise measuring tool would yield $5.8$ m. Suppose as well that this is multiplied: $5.8\cdot 5.8= 33$. Clearly $25 < 33$, but we want to ensure our data it consistently read and therefore a rule is adhered to for some uniformity with which we can rely on. $30$ satisfies the supposition that only one significant figure is being used and $25$ does not.

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  • $\begingroup$ Thanks. 5.8 x 5.8 gives 33.64. $\endgroup$ – user11206 Mar 24 '15 at 17:18

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