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Examples in this article were generated with
4.2.0 by the packages `JuliaCall`

,^{1} `Rmpfr`

,^{2} `microbenchmark`

,^{3} and
`rational,`

^{4} spiced with a little `Python`

^{5} and
`Julia`

^{6} run on a Core i5-8265U @ 1.60GHz (1/4
cores) on Windows 11 build 22000.

- The right-hand badges give the respective section’s ‘level’.

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- These sections are the most important ones. They are – hopefully – easily comprehensible even for novices.

- A somewhat higher knowledge of computer languages (and especially R) is required. May be skipped or reserved for a later reading.

- An advanced knowledge of R is required. Definitely not recommended for beginners.

- If you are not a
**ne**R**d**or computer geek, skipping is recommended. Suggested for experts but might be confusing for others.

- Click to show / hide R code.

Shall we bother about numeric precision?

This article was inspired by a thread at
the BEBA Forum.

It all started with an observation in Excel.

When you enter …

you will get

Well, as expected. But when you enter …

you will get

What? Strange at least.

Both without and *with* parentheses. Amazing!

In all others below it doesn’t matter whether parentheses are used or not.

```
= [0.5, -0.4, -0.1]
x print(sum(x))
# -2.7755575615628914e-17
```

```
= [0.5, -0.4, -0.1]
x # 3-element Vector{Float64}:
# 0.5
# -0.4
# -0.1
print(sum(x))
# -2.7755575615628914e-17
```

What about `R`

?

```
0.5 - 0.4 - 0.1
# [1] -2.775558e-17
```

Gimme more digits, pleeze!

`print(format(0.5 - 0.4 - 0.1, digits = 15))`

`# [1] "-2.77555756156289e-17"`

Root cause analysis:

```
<- matrix(data = c(0.5, -0.4, -0.1,
m 0.5 - 0.4 - 0.1)),
(dimnames = list(c("a", "b", "c",
"sum(a, b, c)"),
"value"))
print(m, digits = 17)
```

```
# value
# a 5.0000000000000000e-01
# b -4.0000000000000002e-01
# c -1.0000000000000001e-01
# sum(a, b, c) -2.7755575615628914e-17
```

Note the ultimate decimal place of `b`

and
`c`

!

```
# Julia
= -0.4;
b typeof(b)
# Float64
= prevfloat(b)
x # -0.4000000000000001
= nextfloat(b)
y # -0.39999999999999997
println(b - x); print(y - a)
# 5.551115123125783e-17
# -5.4
BigFloat(b)
# -0.40000000000000002220446049250313080847263336181640625
"-0.4"
big# -0.4000000000000000000000000000000000000000000000000000000000000000000000000000009
```

Sooner or later we reach the numeric resolution.

```
# Julia
= -0.4;
b = BigFloat(-0.4);
x = big"-0.4";
y println(typeof(b)); println(typeof(x)); print(typeof(y))
# Float64
# BigFloat
# BigFloat
println(x - b); print(y - b)
# 0.0
# 2.220446049250313080847263336181640624999999999999999999999999913638314449055554e-17
= [BigFloat(0.5), BigFloat(-0.4), BigFloat(-0.1)]
x # 3-element Vector{BigFloat}:
# 0.5
# -0.40000000000000002220446049250313080847263336181640625
# -0.1000000000000000055511151231257827021181583404541015625
sum(x)
# -2.77555756156289135105907917022705078125e-17
= [big"0.5", big"-0.4", big"-0.1"]
x # 3-element Vector{BigFloat}:
# 0.5
# -0.4000000000000000000000000000000000000000000000000000000000000000000000000000009
# -0.1000000000000000000000000000000000000000000000000000000000000000000000000000002
sum(x)
# -1.079521069386805578173293982850049946389500045554535173127962933771073975395303e-78
```

However, the ‘higher precision’ is a delusion.

“Reach for the stars, even if you have to stand on a cactus.

Arbitrarily accurate computation with R is
provided by the package `Rmpfr`

.

```
library(Rmpfr)
<- mpfr(c(0.5, -0.4, -0.1), prec = 260)
x
xsum(x)
```

```
# 3 'mpfr' numbers of precision 260 bits
# [1] 0.5
# [2] -0.40000000000000002220446049250313080847263336181640625
# [3] -0.1000000000000000055511151231257827021181583404541015625
# 1 'mpfr' number of precision 260 bits
# [1] -2.77555756156289135105907917022705078125e-17
```

OK, more digits but still not what we expect.

Another example:

```
<- seq(0.40, 0.43, 0.01)
x
xprint(x, digits = 17)
mpfr(x, prec = 260)
```

```
# [1] 0.40 0.41 0.42 0.43
# [1] 0.40000000000000002 0.41000000000000003
# [3] 0.42000000000000004 0.42999999999999999
# 4 'mpfr' numbers of precision 260 bits
# [1] 0.40000000000000002220446049250313080847263336181640625
# [2] 0.41000000000000003108624468950438313186168670654296875
# [3] 0.42000000000000003996802888650563545525074005126953125
# [4] 0.429999999999999993338661852249060757458209991455078125
```

Actually it turned out to be the *most* frequently asked
question about R, the (in)famous
FAQ 7.31.^{7}

What happened here? We have fallen into the trap of floating point arithmetic.

The first attempts to build something we call now a computer^{8} were made
by Charles
Babbage prior to 1840. Of course, both the *Difference
Engine* and the *Analytical
Engine* were – as their names suggest – purely mechanical. The
latter was already programmable (its instruction set developed by Ada Lovelace).
Their numeral system was *decimal* and hence, the examples above
likely would have easily worked.

In 1941 Konrad
Zuse completed construction of the Z3, which was
the first digital computer. Once you deal with electrics (relays, vacuum
tubes), it’s clear why Zuse decided to work with binary digits.

The signal is either off or on .

To convert a decimal number to binary digit, we have to split the number into its integer and fractional part. Procedure for 10.125 as an example.

10 / 2 = 5: r 0

5 / 2 = 2: r 1

2 / 2 = 1: r 0

1 / 2 = 0:
r 1

reordered from least significant bit upwards → [1010]_{2}

0.125 × 2 = 0.25: d 0

0.25 × 2 = 0.50: d 0

0.50 × 2 = 1 : d
1

complete: [10.125]_{10} = [1010.001]_{2}

Now we will see why 0.5–0.4–0.1 is so difficult in the binary system.

[0.5]_{10} = [0.1]_{2}

[0.4]_{10} =
[0.011001100110011001100110011001100110…]_{2} = [0.0110011]_{2}

[0.1]_{10} =
[0.0001100110011001100110011001100110011…]_{2} = [0.00011]_{2}

Only real numbers \(\small{\mathbb{R}\subset2^{\,\mathbb{Z}}}\),
where \(\small{\mathbb{Z}}\) is an
integer {…, –1, 0, 1, …}, can be converted to a binary number without a
remainder. This works for 0.5 (= 2^{–1}) but not for 0.4 and
0.1; we get *periodic* binary numbers, which cannot be stored in
the binary format without truncation.^{9} ^{10}

According to IEEE 754 a binary in double precision holds 64 bits (where 1 bit is the sign, 11 the exponent, and 52 the mantissa). That translates into ~15.7 digits decimal.

```
log(2^52, 10)
abs(log(.Machine$double.eps, 10))
```

```
# [1] 15.65356
# [1] 15.65356
```

Then Ohlbe posted this example, which cought me on the wrong foot first. It boiled down to the question why the second one works (although 5 is not a multiple of two):

```
0.5 - 0.4 - 0.1 == 0
5 - 4 - 1 == 0
```

```
# [1] FALSE
# [1] TRUE
```

[5]_{10} = [101]_{2}

[4]_{10} =
[100]_{2}

[1]_{10} = [1]_{2}

Easy.

I overlooked that the conversion to a binary works for *any*
integer \(\small{\mathbb{Z}}\) within
the range of \(\small{\left\{-2^{-31}=-2,147,483,648\,\ldots\,2^{31}-1=2,147,483,647\right\}}\).

-[2147483648]_{10} =
-[0000000000000000000000000000000]_{2}

[2147483647]_{10} =
+[1111111111111111111111111111111]_{2}

```
<- .Machine$integer.max
a print(a)
print(class(a))
```

```
# [1] 2147483647
# [1] "integer"
```

```
<- as.integer(1)
b <- a + b c
```

`# Warning in a + b: NAs produced by integer overflow`

```
print(c)
class(c)
```

```
# [1] NA
# [1] "integer"
```

Does not work because the integer range is exhausted.

```
<- as.numeric(1) # double precision (float)
d <- a + d
e print(e)
class(e)
```

```
# [1] 2147483648
# [1] "numeric"
```

We get what we expect because a type conversion is performed.

What about the commutative property?

```
<- 0.5
a <- -0.4
b <- -0.1
c cat("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c), "\n")
<- 5
a <- -4
b <- -1
c cat("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c))
```

```
#
# TRUE
# FALSE
# TRUE
#
# TRUE
# TRUE
# TRUE
```

```
# Python
= 0.5
a = -0.4
b = -0.1
c print("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c))
= 5
a = -4
b = -1
c print("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c))
#
# True
# False
# True
#
# True
# True
# True
```

```
# Julia
= 0.5;
a = -0.4;
b = -0.1;
c print("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c), "\n")
#
# true
# false
# true
= 5;
a = -4;
b = -1;
c print("\n", a + b + c == (a + b) + c,
"\n", a + b + c == a + (b + c),
"\n", a + b + c == (a + b + c))
#
# true
# true
# true
```

When dealing with floating point arithmetic, the order (and parentheses) matter (THX to mittyri). No problems with integers.

If you want to compare double precision numbers, say, in the logical
construct of a script, *i.e.*, `if()`

,
`while()`

, `repeat()`

, **do not** use
these goodies:

```
<- 0.5 - 0.4 - 0.1
a <- 0
b == b # most commonly used
a identical(a, b) # not better
```

```
# [1] FALSE
# [1] FALSE
```

BTW, `identical()`

can give unexpected results.

```
<- 2147483647
a <- 2147483647L
b class(a)
class(b)
== b
a identical(a, b)
```

```
# [1] "numeric"
# [1] "integer"
# [1] TRUE
# [1] FALSE
```

Here testing for equality *passes* because the numbers are not
above to maximum possible integer of the system, *i.e.*,
2^{31}–1 (in `R`

that’s
`.Machine$integer.max`

).

However, testing with `identical()`

fails because it compares
not only the *values* but also their *classes*.

Instead use:

```
<- 0.5 - 0.4 - 0.1
a <- 0
b all.equal(a, b)
<- 2147483647
a <- 2147483647L
b all.equal(a, b)
```

```
# [1] TRUE
# [1] TRUE
```

Only this function compares numbers – irrespective of their classes – based on the numeric resolution of a 64 bit double precision numeric, which is \(\small{\approx2.220446\cdot10^{-16}}\). Actually the comparison is performed at its square root or \(\small{\approx1.49011610^{-8}}\).

The source of `all.equal()`

is lengthy but we can mimick
what goes on behind the curtain.

```
<- 2147483647
a <- 2147483647L
b all.equal(a, b)
sqrt(.Machine$double.eps) >= abs(a - b)
```

```
# [1] TRUE
# [1] TRUE
```

We can get \(\small{\pi}\) with up to 16 correct significant digits.

```
cat(formatC(pi, digits = 16, small.mark = "\u2219",
small.interval = 3), "\n")
```

`# 3.141·592·653·589·793`

However, that was a lucky punch because as we have seen above, the 16^{th} is already
inaccurate. But again, don’t dare to ask for more digits. Anything
beyond the 15^{th} significant digit is just ‘noise’.

```
<- "\u2219"
sm cat("64 bit max =", formatC(pi,
digits = 15,
small.mark = sm,
small.interval = 3),
"\n64 bit \u2018noise\u2019 =", formatC(pi,
digits = 31,
small.mark = sm,
small.interval = 3),
"\ncorrect =", paste0("3.141", sm, "592", sm, "653", sm,
"589", sm, "793", sm, "238", sm,
"462", sm, "643", sm, "383", sm,
"279 \u2026"), "\n")
```

```
# 64 bit max = 3.141·592·653·589·79
# 64 bit ‘noise’ = 3.141·592·653·589·793·115·997·963·468·544
# correct = 3.141·592·653·589·793·238·462·643·383·279 …
```

Honestly, I don’t know why it is possible to ask `R`

for more than 15 digits. At least it
should issue a message like

`Please use your wetware before asking!`

We remember from trigonometry that \(\small{\sin\pi=0}\). Given the above, can we really hope for that?

`sin(pi)`

`# [1] 1.224606e-16`

Now we shouldn’t be surprised any more.

```
# Python
import math
print(math.sin(math.pi))
# 1.2246467991473532e-16
```

```
# Julia
sin(pi)
# 1.2246467991473532e-16
sin(BigFloat(pi))
# 1.096917440979352076742130626395698021050758236508687951179005716992142688513354e-77
```

Closer to zero. Will we fare better with `Rmpfr`

?

```
<- Const("pi", prec = 260)
pi. sin(pi.)
```

```
# 1 'mpfr' number of precision 260 bits
# [1] 1.7396371592546498568836643545648074661258190954152778051042783221068713118047497e-79
```

Seems that to hope for zero is futile. However, this ‘better’ result comes with a price, speed.

```
library(microbenchmark)
<- microbenchmark(sin(pi), sin(pi.), times = 1000L)
res options(microbenchmark.unit = "relative")
print(res, signif = 4)
```

```
# Unit: relative
# expr min lq mean median uq max neval cld
# sin(pi) NaN 1.0 1.0 1.0 1.0 1 1000 a
# sin(pi.) Inf 336.9 333.8 336.7 338.7 178 1000 b
```

A funky one^{11} discovered by mittyri:

Not only in Excel…

`1.2e+200 + 1e+100`

`# [1] 1.2e+200`

```
# Python
print(1.2e+200 + 1e+100)
# 1.2e+200
```

```
# Julia
1.2e+200 + 1e+100
# 1.2e200
BigFloat(1.2e+200) + BigFloat(1e+100)
# 1.200000000000000031665409735558622623636694369262012649966820080464248350755499e+200
```

Bad luck (exhausting the double precision). It doesn’t make sense to
hope for the correct

12
000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
000 000 000 000 000 000 000 000 000 000 000 000 000 000 001 000 000 000
000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
000 000 000 000 000 000 000 000 000 000 000 000.

Can you spot the 1?

We learned that we shouldn’t divide by zero. That’s one of the most annoying errors when you start writing own code.

Developers of spreadheets didn’t want to confuse users who never made it beyond basic maths.

An even very small number works but zero doesn’t. But is that correct? Ever came across \(\small{\lim\,f(x)}\)?```
<- data.frame(x = c(0, 1, 1e-10, 1e-250),
df y = 1 / c(0, 1, 1e-10, 1e-250))
$z <- 1 / df$y
dffor (i in 1:nrow(df)) {
$comp[i] <- isTRUE(all.equal(df$x[i], df$z[i]))
df
}names(df)[2:4] <- c("y = 1/x", "z = 1/y", "z == x")
print(df, row.names = FALSE)
```

```
# x y = 1/x z = 1/y z == x
# 0e+00 Inf 0e+00 TRUE
# 1e+00 1e+00 1e+00 TRUE
# 1e-10 1e+10 1e-10 TRUE
# 1e-250 1e+250 1e-250 TRUE
```

Not only \(\small{1/0=\infty}\) but also \(\small{1/\infty=0}\). Nice, though that’s not helpful.

`summary(df[, 1:3])`

```
# x y = 1/x z = 1/y
# Min. :0.00 Min. : 1.0e+00 Min. :0.00
# 1st Qu.:0.00 1st Qu.: 7.5e+09 1st Qu.:0.00
# Median :0.00 Median :5.0e+249 Median :0.00
# Mean :0.25 Mean : Inf Mean :0.25
# 3rd Qu.:0.25 3rd Qu.: Inf 3rd Qu.:0.25
# Max. :1.00 Max. : Inf Max. :1.00
```

```
# Julia
= [0, 1, 1e-10, 1e-250]; y = 1 ./x; z = 1 ./y;
x = [x y z]
a # 4×3 Matrix{Float64}:
# 0.0 Inf 0.0
# 1.0 1.0 1.0
# 1.0e-10 1.0e10 1.0e-10
# 1.0e-250 1.0e250 1.0e-250
:, 3] == a[:, 1]
a[# true
```

What about another infamous candidate, namely \(\small{\log_{e}0}\)?

```
<- data.frame(x = c(exp(1), exp(1) / 1e10,
df exp(1) / 1e250, 0),
y = c(log(exp(1)), log(exp(1) / 1e10),
log(exp(1) / 1e250), log(0)))
$z <- exp(df[, 2])
dffor (i in 1:nrow(df)) {
$comp[i] <- isTRUE(all.equal(df$x[i], df$z[i]))
df
}names(df)[2:4] <- c("y = log(x)", "z = exp(y)", "z == x")
print(df, row.names = FALSE)
```

```
# x y = log(x) z = exp(y) z == x
# 2.718282e+00 1.00000 2.718282e+00 TRUE
# 2.718282e-10 -22.02585 2.718282e-10 TRUE
# 2.718282e-250 -574.64627 2.718282e-250 TRUE
# 0.000000e+00 -Inf 0.000000e+00 TRUE
```

Similarly to the reciprocal of zero above, no fear of infinity: \(\small{\log_{e}0=-\infty}\) and \(\small{\exp(-\infty)=0}\).

```
# Julia
= [exp(1), exp(1)/1e10, exp(1)/1e250, 0]; y = log.(x); z = exp.(y);
x = [x y z]
a # 4×3 Matrix{Float64}:
# 2.71828 1.0 2.71828
# 2.71828e-10 -22.0259 2.71828e-10
# 2.71828e-250 -574.646 2.71828e-250
# 0.0 -Inf 0.0
:, 3] == a[:, 1]
a[# false
2, 3], 3] == a[[2, 3], 1]
a[[# false
1, 4], 3] == a[[1, 4], 1]
a[[# true
```

Interesting. Contrary to `R`

,
the second and third row fail, where the first and fourth pass. Yep, the
last had \(\small{-\infty}\) as an
intermediate result and no problems with \(\small{\exp(-\infty)}\).

For simplicity we can say that \(\small{\log_{e}0}\) is undefined. It would
not be a good idea to trust in a mathematically correct value which
distorts subsequent calculations.

It is reasonable to assume that concentrations \((\small{x \in \mathbb{R}^+)}\) follow a
lognormal distribution. The geometric mean should not work if a value is
zero because it is outside the domain of the lognormal
distribution.

Say, we have an arbitrary long vector of *identical* values and
add a single zero-element to the vector.

```
<- 999
numbers <- 1L
value <- c(rep(value, numbers), 0L)
x <- exp(mean(log(x), na.rm = TRUE))
gm cat(paste0(numbers, " identical values (", value, "), one zero.\n"))
summary(x)
cat("geometric mean:", gm, "\n")
cat("x is", typeof(x),
"\ngeometric mean is", typeof(gm))
```

```
# 999 identical values (1), one zero.
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 0.000 1.000 1.000 0.999 1.000 1.000
# geometric mean: 0
# x is integer
# geometric mean is double
```

```
import StatsBase
= repeat([1], 999); x = push!(x, 0);
x = StatsBase.geomean(x);
gm describe(x)
StatsBase.# Summary Stats:
# Length: 1000
# Missing Count: 0
# Mean: 0.999000
# Minimum: 0.000000
# 1st Quartile: 1.000000
# Median: 1.000000
# 3rd Quartile: 1.000000
# Maximum: 1.000000
# Type: Int64
print("geometric mean: ", gm, "\n", "Type is ", typeof(gm))
# geometric mean: 0.0
# Type is Float64
```

Because: \[\small{\begin{array}{l} x_{i=1\ldots n-1}=1,\;x_{i=n}=0\\ x=\left\{1,\ldots,1,0\right\}\\ \log_{e}x=\left\{0,\ldots,0,-\infty\right\}\\ \overline{\log_{e}x}=\sum \left\{0,\ldots,0,-\infty\right\}/n=-\infty\\ \overline{x}_\textrm{geom.}=\exp(-\infty)=0\;\tiny{\square} \end{array}}\]

Note also the type conversion in `R`

and `Julia`

. Though
`x`

consists of integers, the geometric mean is a double
precision float.

That’s fascinating.

When I copied the output of π to the
clipboard it showed 34 (‼) significant digits 3.141592653589793238462643383279503,
which is correct to the penultimate digit; the last one is rounded up
from 28.
How is that possible? Numbers are represented with a 34 digit mantissa
and an exponent from 10^{−6,413} to 10^{6,144},
*i.e.*, are handled in quadruple
precision (128 bit)!^{13}

Even the original
HP-42S (1988!) represented numbers with a 12 digit mantissa and an
exponent from 10^{−499} to 10^{499}, which is larger
than the IEEE-754 double precision range of 10^{−308} to
10^{308}. I’m impressed.

zero? Sorry, that’s beyond me.

Just to set all of this into perspective:

The diameter of human hair is ≈10^{–12} of the the earth-moon
distance. One Nanometer is ≈10^{–13} of the earth’s equator.
Should we *really* be concerned about an ‘error’ which is three
or more orders of magnitude smaller? 😉

Our measurements are likely never that precise anyway. Furthermore, the
standard defines sophisticiated rounding routines. Hence, in repeated
calculations the error will not propagate upwards.

Hence, the answer to the question

Shall we bother about numeric
precision?

is in general

No.

The physical constant with the highest precision is the Faraday
constant *F*
(9.648 533 212 331 001 84·10^{4} A·s·mol^{–1}) with 18
significant digits. Unless you are an experimental physicist, double
precision numbers are fine. Otherwise, opt for a language supporting
extended precision like GCC
C/C++, Clang, Intel
C++, Object
Pascal, Racket,
Swift,
or get a suitable scientific pocket calculator.

top of section ↩︎ previous section ↩︎

Acknowledgment

Members of the BEBA-Forum: ElMaestro, mittyri, Ohlbe, PharmCat, Shuanghe, and zizou.

Licenses

Helmut Schütz 2022

`R`

and all packages GPL 3.0,
`rational`

, `Free42`

, and `pandoc`

GPL 2.0,
`Python`

Open Source
(GPL compatible), `Julia`

MIT.

1^{st} version March 14, 2021. Rendered June 18, 2022 20:39 CEST
by rmarkdown
via pandoc in 1.85 seconds.

Footnotes and References

Li C, Lai R, Grominski D, Teramo N.

*JuliaCall: Seamless Integration Between R and ‘Julia’.*Package version 0.17.4. 2021-05-14. CRAN.↩︎Maechler M, Heiberger RM, Nash JC, Borchers HW.

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