Class 8: Interval Arithmetic

Methodology of Scientific Research

Andrés Aravena, PhD

March 15, 2023

Dinosaur Bones

Some tourists in the Museum of Natural History are marveling at some dinosaur bones. One of them asks the guard, “Can you tell me how old the dinosaur bones are?”

The guard replies, “They are 3 million, four years, and six months old.”

“That’s an awfully exact number,” says the tourist. “How do you know their age so precisely?”

The guard answers, “Well, the dinosaur bones were three million years old when I started working here, and that was four and a half years ago.”

Being honest

Lets be honest about what we know and what we do not know

We write the values that have real meaning

3 million years means 3±0.5 ⨉ 106

Adding 4.5 years is meaningless

Significant Figures

Which figures are significant?

  • All non-zero digits are significant
    • In 1234 all digits are significant
    • Same in 12.34 and 1.234
  • Zeros surrounded by non-zero are significant
    • Same in 1204. Four significant digits

Which figures are significant ?

  • Zeros to the left are not significant
    • Four significant digits in 0.0001234
  • Zeros to the right are not significant unless there is a decimal point
    • 12340 has four significant digits
    • 123.40 has five significant digits
    • 1234.0 has five significant digits
    • 12340. has five significant digits

Be aware of notation

Notice that 1234.0 is not the same as 1234

Also, 12340. is not the same as 12340

This is a convention but not everybody uses it

It is much safer to use scientific notation

Using scientific notation

It is safer to represent numbers in scientific notation \[ \begin{aligned} 1234.0 & = 1.2340\cdot 10^3\\ 1234 & = 1.234 \cdot 10^3\\ 12340. & = 1.2340 \cdot 10^4\\ 12340 & = 1.234 \cdot 10^4 \end{aligned} \]

All digits in the mantissa are significant

(mantissa is the number being multiplied by 10 to the power of the exponent)

Writing our result

  • Last class we measured the volume of the stone ball
  • Round the uncertainty to a single figure
    • Instead of \(2013.765 ± 78.345\) write \(2013.765 ± 80\)
  • Then round the main value to the last well known place
    • Instead of \(2013.765\) ± 80 write \(2010 ± 80\)
  • The digits “3.765” were a white lie. Let’s not fool ourselves

Interval arithmetic


We will represent measurements as intervals \([x_1, x_2]\)

A binary operation \(\star\) on two intervals is defined by

\[[x_1, x_2] {\,\star\,} [y_1, y_2] = \{ x \star y \, | \, x \in [x_1, x_2] \text{ and } y \in [y_1, y_2] \}.\]

In other words, it is the set of all possible values of \(x \star y\),
where \(x\) and \(y\) are in their corresponding intervals.

Wikipedia: “Interval arithmetic”

Simplification for basic operations

If \(\star\) is either \(+, -, \cdot,\) or \(÷\), then \([x_1, x_2] \star [y_1, y_2]\) is \[ [\min\{ x_1 \star y_1, x_1 \star y_2, x_2 \star y_1, x_2 \star y_2\},\\ \max \{x_1 \star y_1, x_1 \star y_2, x_2 \star y_1, x_2 \star y_2\} ], \] as long as \(x \star y\) is defined for all \(x\in [x_1, x_2]\) and \(y \in [y_1, y_2]\).

Even easier

  • Addition: \[[x_1, x_2] + [y_1, y_2] = [x_1+y_1, x_2+y_2]\]
  • Subtraction: \[[x_1, x_2] - [y_1, y_2] = [x_1-y_2, x_2-y_1]\]


This is the area of a rectangle with varying edges

\([x_1, x_2] \cdot [y_1, y_2]\) is \[[\min \{x_1 y_1,x_1 y_2,x_2 y_1,x_2 y_2\},\\ \max\{x_1 y_1,x_1 y_2,x_2 y_1,x_2 y_2\}]\]

The result interval covers all possible areas, from smallest to the largest

Division needs more attention

\[\frac{[x_1, x_2]}{[y_1, y_2]} = [x_1, x_2] \cdot \frac{1}{[y_1, y_2]},\] where \[\begin{aligned} \frac{1}{[y_1, y_2]} &= \left[\tfrac{1}{y_2}, \tfrac{1}{y_1} \right] \textrm{ if }\;0 \notin [y_1, y_2]\\ \frac{1}{[y_1, 0]} &= \left[-\infty, \tfrac{1}{y_1} \right] \end{aligned}\]


\[\begin{aligned} \frac{1}{[0, y_2]} &= \left [\tfrac{1}{y_2}, \infty \right ] \\ \frac{1}{[y_1, y_2]} &= \left [-\infty, \tfrac{1}{y_1} \right ] \cup \left [\tfrac{1}{y_2}, \infty \right ] \textrm{ if }\;0 \in (y_1, y_2) \end{aligned}\]


  • Exponential function: \(a^{[x_1, x_2]} = [a^{x_1},a^{x_2}]\) for \(a > 1,\)
  • Logarithm: \(\log_a [x_1, x_2] = [\log_a {x_1}, \log_a {x_2}]\) for positive intervals \([x_1, x_2]\) and \(a>1,\)
  • Odd powers: \([x_1, x_2]^n = [x_1^n,x_2^n]\), for odd \(n\in ℕ\)

What happens in even powers?

Density of the stone ball

Measuring the diameter

Let’s measure the diameter using a caliper

It is about 5.3cm

Since the ball is not 100% spherical, the diameter varies between 5.2cm and 5.4cm

Now we can calculate the volume using the formula \[\frac{4}{3} π r^3\]

Calculating the volume

Using the central value

diam <- 5.3
r <- diam/2
[1] 2.65
vol <- 4/3 * pi * r^3
[1] 77.95181

Being pessimistic

Maybe the diameter is 5.2cm. Then

diam_min <- 5.2
r_min <- diam_min/2
vol_min <- 4/3 * pi * r_min^3
[1] 73.62218

Being optimistic

Maybe the diameter is 5.4cm. Then

diam_max <- 5.4
r_max <- diam_max/2
vol_max <- 4/3 * pi * r_max^3
[1] 82.44796

A range of possibilities

The real volume is somewhere in the range

c(vol_min, vol_max)
[1] 73.62218 82.44796

That is, an interval with center

[1] 78.03507

and width

[1] 4.41289

Two possible centers

Notice that

[1] 78.03507

is not the same as

[1] 77.95181

but both values are close (why?)

For now we take the first one

How many decimals?

We can write 78.0350671 ± 4.4128905 cm3

But then, most of the decimals are meaningless

We get a false feeling of precision,
but we really do not know all the decimals

Rounding numbers

First, we round the error term to a single digit

[1] 4

Then we discard all decimals smaller than the error term

[1] 78

so we write 78 ± 4cm3

(we get the same result if we use vol)

Evaluating the density

Each measurement is an interval

We have \[vol=78cm^3 ± 4cm^3\] and \[mass=250gr ± 50gr\]

Now we can calculate the density

We will consider all possible cases


\[\begin{aligned} vol &=[74, 82]cm^3\\ mass&=[200, 300]gr\\ density_{min} &=\min \left\{\frac{200}{74}, \frac{300}{74},\frac{200}{82},\frac{300}{82}\right\}\\ density_{max} &=\max\left\{\frac{200}{74}, \frac{300}{74},\frac{200}{82},\frac{300}{82}\right\}\ \end{aligned} \]


\[ density=[2.4390244, 4.0540541] \frac{gr}{cm^3} \]

Rounding, we get

\[ density=[2.4, 4.1] \frac{gr}{cm^3} \]