SL 1.6—Approximating and estimating

Leading zeros are not significant. The first non-zero digit is 7, so counting of significant figures begins there.

The three significant figures are 7, 4, and 8. The number so far reads $0.0748\dots$

The digit after the 3rd significant figure is 6, which is $\ge 5$, so we round up: $8\to 9$.

$0.074863\approx 0.0749\text{ (to 3 s.f.)}$

We need to round $0.074863$ to 3 significant figures, starting the count at the first non-zero digit.

$0.074$ has only 2 significant figures (7 and 4), so this cannot be the answer to 3 s.f.

$0.075$ has only 2 significant figures (7 and 5). It would be the answer if we were rounding to 2 s.f., not 3.

$0.0748$ is the result of rounding down, but the 4th digit is 6 ($\ge 5$), so we must round up, not down.

$0.0749$ correctly rounds up the 3rd significant figure from 8 to 9, giving 3 significant figures.

The value $26.0$ is given to 1 decimal place, so the half-interval is $0.5\times 10^{-1}=0.05$.

$\text{Lower bound}=26.0-0.05=25.95$

$\text{Upper bound}=26.0+0.05=26.05$

The lower bound uses $\le$ and the upper bound uses $<$ (strict), because a value exactly equal to 26.05 would round up to 26.1, not down to 26.0. So: $25.95\le L<26.05$

We need the correct inequality notation for bounds of $26.0$ rounded to 1 d.p., including the correct inequality signs.

$25.5\le L<26.5$ uses a half-interval of 0.5, which would be correct for rounding to the nearest whole number — not to 1 decimal place.

$25.9\le L<26.1$ uses a half-interval of 0.1, which is the full unit at 1 d.p., not half of it. The half-interval should be 0.05.

$25.95<L\le 26.05$ has the inequality signs reversed. The lower bound must use $\le$ and the upper bound must use $<$ by mathematical convention.

$25.95\le L<26.05$ has the correct half-interval of 0.05 and the correct inequality signs.