nyuuzyou/wikis huggingface dataset:
Number of Primes is Infinite/Proof 6

== Theorem ==
== Proof ==
Let $p_1, p_2, \ldots, p_j$ be the first $j$ primes.
For each real $x$ and natural number $j$, let:
:$\map {M_j} x = \set {n \in \N \mid n \le x, \, n \text { is not divisible by any prime } p \text { with } p > p_j}$
Define:
:$\map {N_j} x = \# \map {M_j} x$
Let $n \in \map {M_j} x$ for some $x$, $j$. 
We can write: 
:$n = n_*^2 m$
where $n_*$ and $m$ are integers with $m$ square-free.
By the Fundamental Theorem of Arithmetic, we may uniquely write: 
:$\ds m = \prod_{i \mathop = 1}^j p_j^{e_j}$
where $e_j \in \set {0, 1}$.
Since there are $2$ choices for each $e_j$, there are $2^j$ possible values of $m$.
Note that since $m \ge 1$, we have: 
:$n_*^2 \le n \le x$
so:
:$n_* \le \sqrt n \le \sqrt x$
So, there at most $\sqrt x$ possible values of $n_*$. 
So, there are at most $2^j \sqrt x$ possible values of $n$.
That is:
:$\map {N_j} x \le 2^j \sqrt x$
 that there are only finitely many primes.
In particular, assume that there are $k$ primes. 
Then:
:$\map {N_k} x = x$
for each $x$, since there are no primes exceeding $p_k$. 
So, we have: 
:$x \le 2^k \sqrt x$
for all $x$. 
But this does not hold for all $x$, for example taking $x = 2^{2 k} + 1$. 
So, we have arrived at a contradiction.
So there are infinitely many primes.
Category:Euclid's Theorem
Number of Primes of Form n^2 + 1

== Conjecture ==
Let $\map P x$ denote the number of prime numbers of the form $n^2 + 1$ less than $x$.
Then:
:$\map P x \sim C \dfrac {\sqrt x} {\ln x}$
where:
== Historical Note ==
== Sources ==
* : $1,3727 \ldots$
Category:Prime Numbers
Category:Unproven Hypotheses
Number of Primes up to n Approximates to Eulerian Logarithmic Integral

== Theorem ==
The prime-counting function approximates to the Eulerian logarithmic integral:
== Proof ==
== Also see ==
* Prime Number Theorem
== Sources ==
* : $10^{10^{10^{34}}}$
* : $10^{10^{10^{34}}}$
* 
Category:Prime Number Theorem
Number of Primes up to n Approximates to Eulerian Logarithmic Integral/Mistake

== Source Work ==
:
:The Dictionary
::$10^{10^{10^{34}}}$
:
:The Dictionary
::$10^{10^{10^{34}}}$
== Mistake ==
:The number of primes less than or equal to $n$ is approximately $\ds \int_0^n \frac {\d x} {\log x}$.
== Correction ==
That should be $\ds \int_2^n \frac {\d x} {\log x}$.
$\log x$ in this instance denotes the natural logarithm function.
== Sources ==
* : $10^{10^{10^{34}}}$
* : $10^{10^{10^{34}}}$
Category:Mistakes/Curious and Interesting Numbers
Number of Quadratic Residues of Prime

== Theorem ==
Let $p$ be an odd prime.
Then $p$ has $\dfrac {p - 1} 2$ quadratic residues and $\dfrac {p - 1} 2$ quadratic non-residues.
The quadratic residues are congruent modulo $p$ to the integers $1^2, 2^2, \ldots, \paren {\dfrac {p - 1} 2}$.
== Proof ==
The quadratic residues of $p$ are the integers which result from the evaluation of the squares:
:$1^2, 2^2, \ldots, \paren {p - 1}^2$ modulo $p$
But:
:$r^2 = \paren {-r}^2$
and so these $p - 1$ integers fall into congruent pairs modulo $p$, namely:
Therefore each quadratic residue of $p$ is congruent modulo $p$ to one of the $\dfrac {p - 1} 2$ integers $1^2, 2^2, \ldots, \paren {\dfrac {p - 1} 2}^2$.
Note that as $r^2 \not \equiv 0 \pmod p$ for $1 \le r < p$, the integer $0$ is not among these.
All we need to do now is show that no two of these integers are congruent modulo $p$.
So, suppose that $r^2 \equiv s^2 \pmod p$ for some $1 \le r \le s \le \dfrac {p - 1} 2$.
What we are going to do is prove that $r = s$.
Now $r^2 \equiv s^2 \pmod p$ means that $p$ is a divisor of $r^2 - s^2 = \paren {r + s} \paren {r - s}$.
From Euclid's Lemma either:
:$p \divides \paren {r + s}$
or:
:$p \divides \paren {r - s}$
$p \divides \paren {r + s}$ is impossible as $2 \le r + s \le p - 1$.
Take $p \divides \paren {r - s}$.
As $0 \le r - s < \dfrac {p - 1} 2$, that can happen only when:
:$r - s = 0$
or:
:$r = s$
So there must be exactly $\dfrac {p - 1} 2$ quadratic residues.
That means there must also be exactly $\dfrac {p - 1} 2$ quadratic non-residues.
Category:Prime Numbers
Category:Quadratic Residues
Number of Random Fractional Reals whose Total Exceeds 1

== Theorem ==
Let real numbers be selected at random following a continuous uniform distribution from the interval $\closedint 0 1$ until their total sum is greater than $1$.
The expectation of the number of selections is Euler's number $e$.
== Proof ==
== Sources ==
* 
* : $2 \cdotp 71828 \, 18284 \, 59045 \, 23536 \, 02874 \, 71352 \, 66249 \, 77572 \, 47093 \, 69995 \ \ldots$
Category:Euler's Number
Category:Probability Theory
Number of Regions by dividing Circle by Chords

== Theorem ==
Let $n$ points be marked on the circumference of a circle $C$.
Let chords be drawn between each pair of these points.
For each $n$, the maximum number $\map C n$ of regions into which $C$ can be divided is as follows:
:
$n$ 	$\map C n$
$1$ 	$1$
$2$ 	$2$
$3$ 	$4$
$4$ 	$8$
$5$ 	$16$
$6$ 	$31$
$7$ 	$57$
$8$ 	$99$
$9$ 	$163$
$10$	$256$
== Proof ==
: $\quad$ $\quad$ 
: $\quad$ $\quad$ 
== Sources ==
* : $31$
Category:Geometry
Number of Regions in Plane Defined by Given Number of Lines

== Theorem ==
The maximum number $L_n$ of regions in the plane that can be defined by $n$ straight lines in the plane is:
:$L_n = \dfrac {n \paren {n + 1} } 2 + 1$
== Proof ==
=== Setting up a Recurrence Rule ===
First we consider the plane with no lines at all.
This has one region, so $L_0 = 1$.
Now when we have one line, we divide the plane into two regions, so $L_1 = 2$.
Now consider the $n$th line.
This increase the number of regions by $k$ iff it splits $k$ of the old regions.
It can split $k$ of the old regions it hits the existing lines on the plane in $k-1$ places.
Two straight lines can intersect in at most one point.
So the new line can intersect the $n - 1$ old ones in at most $n - 1$ different points.
Therefore $k \le n$.
So we see that $L_n \le L_{n - 1} + n$.
Now, it is always possible to place the $n$th line so that:
:$(1): \quad$ It is not parallel to any of the others, and therefore intersects all the other $n - 1$ lines
:$(2): \quad$ It does not go through any of the existing intersection points (so intersects them all in different places).
Thus we see that $L_n \ge L_{n - 1} + n$.
Hence the recurrence:
:$L_n = L_{n - 1} + n$
=== Solution of Recurrence ===
Using induction, we show that:
:$L_n = \dfrac {n \paren {n + 1} } 2 + 1$
The base case is straightforward:
:$L_0 = 1 = \dfrac {0 \paren {0 + 1} } 2 + 1$
:$L_1 = 2 = \dfrac {1 \paren {1 + 1} } 2 + 1$
Now assume the induction hypothesis:
:$L_k = \dfrac {k \paren {k + 1} } 2 + 1$
and try to show:
:$L_{k + 1} = \dfrac {\paren {k + 1} \paren {k + 2} } 2 + 1$
Hence the induction step:
Hence the result by induction.
== Examples ==
== Historical Note ==
== Sources ==
* 
* : $22$
* $\S 1.2$
* : $22$
Category:Discrete Mathematics
Category:Number of Regions in Plane Defined by Given Number of Lines
Number of Regions in Plane Defined by Given Number of Lines/Examples/6

== Example of Number of Regions in Plane Defined by Given Number of Lines ==
With $6$ lines, the plane can be divided into a maximum of $22$ regions:
:
== Proof ==
From Number of Regions in Plane Defined by Given Number of Lines, the maximum number $L_n$ of regions in the plane that can be defined by $n$ straight lines in the plane is:
:$L_n = \dfrac {n \paren {n + 1} } 2 + 1$
Here $n = 6$, so:
== Sources ==
* : $22$
* : $22$
Category:Number of Regions in Plane Defined by Given Number of Lines
Category:22
Number of Regular Stellated Odd n-gons

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive odd integer.
Then there are $\dfrac {n - 1} 2$ distinct regular stellated $n$-gons.
== Proof ==
Let $P$ be a regular stellated $n$-gons.
Let the $n$ vertices of $P$ be $p_1, p_2, \dotsc, p_n$.
These will be arranged on the circumference of a circle $C$, dividing $C$ into $n$ arcs of equal length.
Once we have chosen the first side of $P$, the others are all the same length and are completely determined by that first side.
, the first vertex of $P$ is chosen to be $p_1$
We can choose that first side as follows:
:$p_1 p_2, p_1 n_3, \ldots, p_1 p_{n - 1}, p_1 p_n$
But we have that:
:$\size {p_1 p_2} = \size {p_1 p_n}$
:$\size {p_1 p_3} = \size {p_1 p_{n - 1} }$
and so on, down to:
:$\size {p_1 p_{\paren {n - 1} / 2} }= \size {p_1 p_{\paren {n + 1} / 2} }$
where $\size {p_a p_b}$ denotes the length of the line $p_1 p_b$.
So for the $n - 1$ lines that are chosen for the first side of $P$, each is paired with another of the same length.
Hence there are $\dfrac {n - 1} 2$ ways of choosing the first side of $P$.
Thus there are $\dfrac {n - 1} 2$ distinct regular stellated $n$-gons.
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$
Category:Regular Stellated Polygons
Number of Representations as Sum of Two Primes

== Theorem ==
The number of ways an integer $n$ can be represented as the sum of two primes is no greater than the number of primes in the interval $\closedint {\dfrac n 2} {n - 2}$.
== Proof ==
Let $n = p + q$ where $p \le q$ and both $p$ and $q$ are primes.
There can be no more different representations of $n$ as $p + q$ than there are the number of possible options for $q$.
As $q \ge p$, it follows that $q \ge \dfrac n 2$.
Note that as $p \ge 2$, it follows that $q \le n - 2$.
The number of possible values of $q$ is therefore equal to the number of primes between $\dfrac n 2$ and $n - 2$ inclusive.
That is, the number of primes in the interval $\closedint {\dfrac n 2} {n - 2}$.
Hence the result.
== Sources ==
* 
* : $210$
Category:Prime Numbers
Number of Selections from n Objects of 2 Types

== Theorem ==
Let $S$ be a collection of $n$ objects, consisting of:
:$p$ objects of one type
:$q$ objects of another type.
The total number $N$ of different non-empty selections from $S$ is given by:
:$N = 2^n \paren {p + 1} \paren {q + 1} - 1$
== Proof ==
For each object, we have $2$ choices: whether to select or whether not.
From Cardinality of Power Set of Finite Set, this gives us $2^n$ options.
From the $p$ things of the first type, we may take $0, 1, 2, \ldots, p$ objects.
Hence we have $\paren {p + 1}$ choices.
Similarly, from the $q$ things of the second type, we have $\paren {q + 1}$ choices.
Hence we have $\paren {p + 1}$ choices.
Because the non-empty selection is excluded, this gives:
:$N = 2^n \paren {p + 1} \paren {q + 1} - 1$
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of selections from $n$ different things, $p$ similar things of one kind and $q$ similar things of another kind, if any number may be taken
Category:Number of Arrangements of n Objects of m Types
Number of Selections of 1 or More from Set

== Theorem ==
Let $S$ be a set of $n$ objects.
The total number $N$ of ways that at least $1$ object can be selected from $S$ is:
:$N = 2^n - 1$
== Proof ==
This is equivalent to counting the number of non-empty subsets of $S$.
From Cardinality of Power Set of Finite Set, the total number of subsets of $S$ is $2^n$.
This includes the empty set.
Excluding the empty set from the count gives the result.
== Examples ==
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of selections from $n$ things if any number may be taken
Category:Number of Selections of 1 or More from Set
Category:Combinations
Number of Selections of 1 or More from Set/Examples/6

== Example of Use of Number of Selections of $1$ or More from Set ==
Let there be $6$ different flowers in a vase.
Then there are $63$ different ways of selecting at least one of these flowers.
== Proof ==
Let $S$ be the set of flowers.
Let $N$ be the number of different ways of selecting at least one of these flowers.
We have:
:$\card S = 6$
where $\card S$ denotes the cardinality of $S$.
Hence from Number of Selections of $1$ or More from Set:
:$N = 2^6 - 1 = 63$
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $9$
Category:Number of Selections of 1 or More from Set
Number of Set Partitions by Number of Components

== Theorem ==
Let $S$ be a (finite) set whose cardinality is $n$.
Let $\map f {n, k}$ denote the number of different ways $S$ can be partitioned into $k$ components.
Then:
:$\ds \map f {n, k} = {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \map f {n, k} = {n \brace k}$
$\map P 0$ is the degenerate case:
:$\ds \map f {0, k} = \delta_{0 k} = {0 \brace k}$
That is: the empty set can be partitioned one and only one way: into $0$ subsets.
Thus $\map P 0$ is seen to hold.
The remainder of the proof considers $n \in \Z_{> 0}$.
First we note that when $k < 1$ or $k > n$:
:$\ds \map f {n, k} = 0 = {n \brace k}$
Hence, throughout, we consider only such $k$ as $1 \le k \le n$.
We define the representative set of cardinality $n$ to be:
:$S_n := \set {1, 2, \ldots, n}$
=== Basis for the Induction ===
$\map P 1$ is the case $\map f {1, 1}$.
There is exactly one way to partition $\set 1$, and that is:
:$\set {\set 1}$
From Stirling Number of the Second Kind of Number with Self:
:$\ds {1 \brace 1} = 1$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
:$\ds \map f {m, k} = {m \brace k}$
from which it is to be shown that:
:$\ds \map f {m + 1, k} = {m + 1 \brace k}$
=== Induction Step ===
This is the induction step:
By definition, the number of partitions of $S_m$ into $k$ subsets is $\map f {m, k}$.
A partition of $S_{m + 1}$ can be generated by adding element $m + 1$ into one of the existing partitions of $S_m$.
There are two ways this can be done:
:$(1): \quad$ The subset $\set {m + 1}$ may be added, in one way, to one of the partitions of $S_m$ into $k - 1$ subsets.
:$(2): \quad$ The element $m + 1$ may be added to any one of the $k$ subsets in one of the partitions of $S_m$ into $k$ subsets.
Option $(1)$ gives $1$ partition of $S_{m + 1}$ for each partition of $S_m$ into $k - 1$ subsets, that is: $\map f {m, k - 1}$.
Option $(2)$ gives $k$ partitions of $S_{m + 1}$ for each partition of $S_m$ into $k$ subsets, that is: $k \map f {m, k}$.
Thus:
:$\map f {m + 1, k} = k \map f {m, k} + \map f {m, k - 1}$
By the induction hypothesis:
:$\ds \map f {m + 1, k} = k {m \brace k} + {m \brace k - 1}$
So by definition of Stirling numbers of the second kind:
:$\ds \map f {m + 1, k} = {m + 1 \brace k}$
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n, k \in \Z_{\ge 0}: \map f {n, k} = {n \brace k}$
== Examples ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $64$
Category:Combinatorics
Category:Stirling Numbers
Category:Number of Set Partitions by Number of Components
Category:Set Partitions
Number of Set Partitions by Number of Components/Examples/4 into 2

== Example of Number of Set Partitions by Number of Components ==
A set with $4$ elements $\set {1, 2, 3, 4}$ can be partitioned into $2$ subsets in $\ds {4 \brace 2} = 7$ ways:
Let $T$ be the set with $3$ elements $\set {1, 2, 3}$.
The partition of $T$ into $1$ set is:
:$\set {1, 2, 3}$
to which $\set 4$ can be added in one way to achieve:
:$\set {1, 2, 3 \mid 4}$
The partitions of $T$ into $2$ sets are:
:$\set {1, 2 \mid 3}$
:$\set {1, 3 \mid 2}$
:$\set {2, 3 \mid 1}$
to which the element $4$ can be added in $2$ ways each:
:$\set {1, 2, 4 \mid 3}$
:$\set {1, 2 \mid 3, 4}$
:$\set {1, 3, 4 \mid 2}$
:$\set {1, 3 \mid 2, 4}$
:$\set {2, 3, 4 \mid 1}$
:$\set {2, 3 \mid 1, 4}$
Thus we have a total of $7$ ways of partitioning a $4$-element set into $2$ components.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $64$
Category:Number of Set Partitions by Number of Components
Category:Examples of Set Partitions
Number of Significant Figures in Result of Addition or Subtraction

== Theorem ==
Let $m$ and $n$ be numbers.
Let $d_m$ and $d_n$ be the position of the of $m$ and $n$ respectively.
Then the in either $m + n$ or $m - n$ is in the position corresponding to the greater significant digit of $d_m$ and $d_n$.
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Computations
Category:Significant Figures
Number of Significant Figures in Result of Addition or Subtraction/Examples

== Examples of Use of Number of Significant Figures in Result of Addition or Subtraction ==
=== Significant Figures of $3 \cdotp 16 + 2 \cdotp 7$ ===
=== Significant Figures of $83 \cdotp 42 - 72$ ===
=== Significant Figures of $47 \cdotp 816 - 25$ ===
Category:Examples of Significant Figures
Number of Significant Figures in Result of Addition or Subtraction/Examples/3.16 + 2.7

== Example of Use of Number of Significant Figures in Result of Addition or Subtraction ==
:$3 \cdotp 16 + 2 \cdotp 7 = 5 \cdotp 9$
== Proof ==
We have that:
:the leftmost significant figure $d_m$ of $3 \cdotp 16$ is the $10^{-2}$ position
:the leftmost significant figure $d_n$ of $2 \cdotp 7$ is the $10^{-1}$ position.
So from Number of Significant Figures in Result of Addition or Subtraction:
::the leftmost significant figure of $3 \cdotp 16 + 2 \cdotp 7$ is the rightmost of $10^{-2}$ and $10^{-1}$, that is $10^{-1}$.
== Sources ==
* : Chapter $1$: Computations: Example 1.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Addition or Subtraction/Examples/47.816 - 25

== Example of Use of Number of Significant Figures in Result of Addition or Subtraction ==
:$47 \cdotp 816 - 25 = 22 \cdotp 816$
on the assumption that $25$ is exact.
== Proof ==
We have that:
:the $d_m$ of $47 \cdotp 816$ is the $10^{-3}$ position
:the $d_n$ of $25$ is the $10^{-\infty}$ position.
So from Number of Significant Figures in Result of Addition or Subtraction:
::the of $47 \cdotp 816 - 25$ is the greater significant digit of $10^{-3}$ and $10^{-\infty}$, that is $10^{-3}$.
== Sources ==
* : Chapter $1$: Computations: Example 3.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Addition or Subtraction/Examples/83.42 - 72

== Example of Use of Number of Significant Figures in Result of Addition or Subtraction ==
:$83 \cdotp 42 - 72 = 11$
== Proof ==
We have that:
:the $d_m$ of $83 \cdotp 42$ is the $10^{-2}$ position
:the $d_n$ of $72$ is the $10^0$ position.
So from Number of Significant Figures in Result of Addition or Subtraction:
::the of $83 \cdotp 42 - 72$ is the greater significant digit of $10^{-2}$ and $10^0$, that is $10^0$.
== Sources ==
* : Chapter $1$: Computations: Example 2.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Division

== Theorem ==
Let $m$ and $n$ be numbers which are presented to $d_m$ and $d_n$ significant figures respectively.
Then the most significant figures that $\dfrac m n$ can have is $\min \set {d_m, d_n}$.
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Computations
Category:Significant Figures
Number of Significant Figures in Result of Division/Examples/1.648 over 0.023

== Example of Use of Number of Significant Figures in Result of Division ==
:$\dfrac {1 \cdotp 648} {0 \cdotp 023} = 72$
== Proof ==
We have that:
:the number of significant figures $d_m$ in $1 \cdotp 648$ is $4$
:the number of significant figures $d_n$ in $0 \cdotp 023$ is $2$
So from Number of Significant Figures in Result of Division:
:the number of significant figures in $\dfrac {1 \cdotp 648} {0 \cdotp 023}$ can be no more than $\min \set {4, 2}$, that is $2$.
== Sources ==
* : Chapter $1$: Computations: Example 2.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Multiplication

== Theorem ==
Let $m$ and $n$ be numbers which are presented to $d_m$ and $d_n$ significant figures respectively.
Then the most significant figures that $m \times n$ can have is $\min \set {d_m, d_n}$.
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Computations
Category:Significant Figures
Number of Significant Figures in Result of Multiplication/Examples/73.24 x 4.52

== Example of Use of Number of Significant Figures in Result of Multiplication ==
:$73 \cdotp 24 \times 4 \cdotp 52 = 331$
== Proof ==
We have that:
:the number of significant figures $d_m$ in $73 \cdotp 24$ is $4$
:the number of significant figures $d_n$ in $4 \cdotp 52$ is $3$
So from Number of Significant Figures in Result of Multiplication:
:the number of significant figures in $73 \cdotp 24 \times 4 \cdotp 52$ can be no more than $\min \set {3, 4}$, that is $3$.
== Sources ==
* : Chapter $1$: Computations: Example 1.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Multiplication/Examples/8.416 x 50

== Example of Use of Number of Significant Figures in Result of Multiplication ==
:$8 \cdotp 416 \times 50 = 420 \cdotp 8$
on the assumption that $50$ is exact.
== Proof ==
We have that:
:the number of significant figures $d_m$ in $8 \cdotp 416$ is $4$
:the number of significant figures $d_n$ in $50$ is unlimited
So from Number of Significant Figures in Result of Multiplication:
:the number of significant figures in $8 \cdotp 416 \times 50$ can be no more than $\min \set {4, \infty}$, that is $4$.
== Sources ==
* : Chapter $1$: Computations: Example 4.
Category:Examples of Significant Figures
Number of Significant Figures in Result of Square Root/Examples/Root of 38.7

== Example of Use of Number of Significant Figures in Result of Square Root ==
:$\sqrt {38 \cdotp 7} = 6\cdotp 22$
== Proof ==
We have that:
:the number of significant figures in $38 \cdotp 7$ is $3$
So from Number of Significant Figures in Result of Square Root:
:the number of significant figures in $\sqrt {38 \cdotp 7}$ can be no more than $3$.
== Sources ==
* : Chapter $1$: Computations: Example 3.
Category:Examples of Significant Figures