Greatest Common Divisor

S2: Divisibility

For the purposes of the following definitions and lemmas, $U = Z$ (including negatives):

Divisibility

$a | b$ (when $a \neq 0$ ) if $\exists x$ such that $b = a x$ . We say $a$ divides $b$ . If $a$ cannot divide $b$ then we write $a ∤ b$ .

If $0 < a < b$ then $a$ is a proper divisor of $b$ . It is best understood that $a \neq 0$ for (pretty much) all of these cases. Further:

$0 | x$ is always FALSE (undefined when $x = 0$ )
$x | 0$ is always TRUE (for all $x$ )

One interesting thing is that the notation $a^{K} | | b$ is sometimes used to indicate $a^{K} | b \land a^{K + 1} ∤ b$ .

Properties of Divisibility

$a | b$ implies $a | b c$ $\forall c$
$a | b$ and $b | c$ implies $a | c$
$a | b$ and $a | c$ implies $a | (b x + c y)$ for all $x, y$
$a | b$ and $b | a$ implies $a = \pm b$
$a | b$ when $a, b > 0$ implies $a \leq b$
If $m \neq 0$ then $a | b$ iff $m a | m b$

Proof
(1): If $a | b$ then $\exists x$ where $b = a x$ . Let $c$ be arbitrary. Then multiply both sides to get $b c = a x c$ . Let $x^{'} = x c$ . Then $b c = a x^{'}$ so by definition then $a | b c$ as required.

(2): If $a | b$ and $b | c$ then $\exists x, y$ where $b = a x$ and $c = b y$ . Substitute the first into second equation to get $c = (a x) y = a (x y)$ . Since $x y$ is an integer then by definition then $a | c$ .

(3): If we prove that $a | b, a | c$ implies $a | (b + c)$ then we are done (just use (1) after that). Let $a | b, a | c$ , so then $b = a x$ and $c = a y$ . Add both together to get $b + c = a (x + y)$ . Clearly $x + y$ is an integer so then $a | (b + c)$ .

(4): If $a | b, b | a$ then $b = a x$ and $a = b y$ . Substitute the left into the right to get that $a = (a x) y = a (x y)$ . Dividing by $a$ gives that $1 = x y$ . The only two cases are that $x, y = 1$ or $x, y = - 1$ . If $x, y = 1$ then $a = b (1) = b$ and in the other case $a = b (- 1) = - b$ so in either case we have $a = \pm b$ as expected.

(5): If $a | b$ and $a, b > 0$ then $\exists x$ where $b = a x$ . Since $b > 0$ and $a > 0$ we require that $x > 0$ . We can therefore use induction over $x$ :

If $x = 1$ then $a = b$ satisfied $a = a x \leq b$ .
Suppose the $x$ case has it that $a \leq a x \leq b$ . If we consider $x + 1$ then:
$b = a (x + 1) \geq a x \geq a$
In all cases we have $a \leq b$ as expected.

(6): Suppose $m \neq 0$ . $a | b$ iff $\exists x$ where $b = a x$ iff $m b = m a x$ iff $(m b) = (m a) x$ iff $m b | m a$ .
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Notice that this extends for all linear combinations of integers too:
$a | b_{1}, a | b_{2}, . . ., a | b_{n} \Rightarrow a | \sum_{j = 1}^{n} b_{j} x_{j} \forall x_{j} \in Z$
Property 2 can be extended similarly:
$a_{1} | a_{2}, a_{2} | a_{3}, . . ., a_{n - 1} | a_{n} \Rightarrow a_{1} | a_{n}$

Using these properties:

$6 | 18$ so with $c = 2$ that implies that $6 | 36$ . Multiplication can only add factors.
$6 | 18$ and $18 | 180$ so then $6 | 180$ . A 'divisibility chain' can only really add factors the farther you go.
Comes from the fact that factors distribute on addition. If $6 | 12$ and $6 | 6$ then $6 | 12 + 6$ because we can take the same factor from both and 'divide' them out.
If $a | b$ and $b | a$ then $\frac{b}{a}$ and $\frac{a}{b}$ are integers, so clearly they have to equal $\pm 1$ .
If $a | b$ then when $a, b > 0$ then $\frac{b}{a} \geq 0$ so clearly $a \leq b$ as a result (when $a \neq 0$ )
Comes from just multiplying fractions on the top and bottom by the same factor.

Division Algorithm

Given any integers $a, b \in Z$ with $a > 0$ then $! \exists q, r \in Z$ such that:
$b = q a + r$
where $0 \leq r < | a |$ . In the case $a ∤ b$ then $r$ satisfies the strong inequalities $0 < r < | a |$ .

Proof
Consider the arithmetic progression:
$\dots, b - 3 a, b - 2 a, b - a, b, b + a, b + 2 a, b + 3 a, \dots$
In this sequence, select the smallest non-negative member and denote it by $r$ . Thus by definition $r$ satisfies the inequalities of the theorem. But also $r$ being in the sequence is of the form $b - q a$ and thus $q$ is defined in terms of $r$ .

To show uniqueness, suppose $q_{1}, r_{1}$ are other integers that satisfy the algorithm. First we prove $r_{1} = r$ . Suppose not, so then we can say that $r < r_{1}$ (WLOG) so then $0 < r_{1} - r < a$ . See that $r_{1} - r = a (q - q_{1})$ (by plugging in $b = q_{1} a + r_{1} = q a + r$ , so then $a | (r_{1} - r)$ . This contradicts Properties of Divisibility, since then (using property 3) we have $a | r_{1}$ and $a | r$ which is impossible by the inequalities we imposed. Thus $r = r_{1}$ and consequently $q = q_{1}$ by plugging into the DA equation.

For the consideration when $a < 0$ , just restate the theorem to use $| a |$ instead, and we get similar results.
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Why call it an algorithm?

Even though the theorem above shows an algorithm to find $q, r$ , something like a computer would do the arithmetic by hand to find both. Really, the theorem helps show the existence of $q, r$ and their uniqueness, rather than a 'good' algorithm for it.

A good way to get these number is just to divide normally, then get the quotient and find $r = b - q a$ :
$\frac{61}{13} = ? \Rightarrow 13 \overset{―}{) 61} = 4 R9 \Rightarrow q = 4, r = 9 \Rightarrow 61 = 13 \cdot 4 + 9$

Greatest Common Divisor (GCD)

$a$ is a common divisor of $b$ and $c$ in case $a | b$ and $a | c$ .

There are only a finite number of divisors of any non-zero integer, so there is only a finite number of common divisors of $b, c$ except in the case $b = c = 0$ . If at least one of $b$ and $c$ is not $0$ , the greatest among their common divisors is called the greatest common divisor of $b$ and $c$ and is denoted either $(b, c)$ or $gcd (b, c)$ . Similarly, we denote the greatest common divisor $g$ of the integers $b_{1}, . . ., b_{n}$ not all zero by $(b_{1}, . . ., b_{n}) = gcd (b_{1}, . . ., b_{n})$ .

We note that $gcd (b, c) \geq 1$ for all $b, c \neq 0$ .

Linear combination of GCD

If $g = gcd (b, c)$ then $\exists x_{0}, y_{0}$ such that $g = gcd (b, c) = b x_{0} + c y_{0}$

Proof
Consider the linear combinations $b x + c y$ , where $x, y$ range over all the integers. Clearly then:
$S = {b x + c y | x, y \in Z} = {\dots, - x - y, - x, - x + y, \dots, - y, 0, y, \dots, x - y, x, x + y, \dots}$
Clearly $0$ is in the set with $x = y = 0$ . Now, choose $x_{0}, y_{0}$ such that $b x_{0} + c y_{0}$ is the least positive integer $l$ in this set (we can do this by the well-ordering principle). Thus $l = b x_{0} + c y_{0}$ .

Next we prove $l | b, l | c$ . Doing the first shows the second by analogy. Proving $l | b$ , by contradiction assume that $l ∤ b$ . Using our Division Algorithm then $! \exists q, r$ such that $b = l q + r$ where $0 < r < l$ (notice the stricter inequality here). Therefore then we have $r = b - l q = b - q (b x_{0} + c y_{0}) = b (1 - q x_{0}) + c (- q y_{0})$ . Notice the form of the equation, it allows us to put $r \in S$ . But this contradicts the fact that $l$ is the least positive integer in the set $S$ (since $r$ would be the least positive integer instead). By contradiction then $l | b$ . Similarly, $l | c$ .

Since $g$ is the greatest common divisor of $b, c$ we may write $b = g B$ and $c = g C$ and:
$l = b x_{0} + c y_{0} = g B x_{0} + g C y_{0} = g (B x_{0} + C y_{0})$
Thus clearly then $g | l$ and so by part 5 of Property 5 then $g \leq l$ . Now $g < l$ is impossible since $g$ is the greatest common divisor, so clearly $g = l = b x_{0} + c y_{0}$ .
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Using Linear combination of GCD

For instance, if we have it that $gcd (180, 282) = 6$ then $6 = 180 x_{0} + 282 y_{0}$ has the solution $x_{0} = 11$ and $y_{0} = - 7$ :
$6 = 180 (11) + 282 (- 7) = 1980 - 1974$
Notice the proof itself doesn't really show how to find these $x_{0}, y_{0}$ values. We'll see that later in Euclid's Algorithm.

Properties of the GCD

The GCD $g = gcd (b, c)$ can be characterized as:

$g$ is the least positive value of $b x + c y$ where $x, y$ range over all $Z$
It is the positive common divisor of $b$ and $c$ that is divisible by every common divisor.

Proof
(1): Use our linear combination argument, this is the smallest by the argument for the proof (we used the smallest element).

(2): If $d$ is any common divisor of $b$ and $c$ then $d | g$ by Property 3 of the properties of divisibility. Property 4 further restricts only one set of integers with this property.
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Clearly if any divisor $d$ of the form $d = b x + c y$ then $d$ is not necessarily $gcd (b, c)$ (ie: the converse isn't necessarily true).

Generalization of Linear Combinations of GCD

Given integers $b_{1}, . . ., b_{n}$ , not all zero, with greatest common divisor $g$ , $\exists x_{1}, . . ., x_{n} \in Z$ such that:
$g = gcd (b_{1}, . . ., b_{n}) = \sum_{j = 1}^{n} b_{j} x_{j}$
Furthermore $g$ is the least positive value of the linear form $\sum_{j = 1}^{n} b_{j} y_{j}$ where all $y_{j}$ range over $Z$ . Also $g$ is the positive common divisor of $b_{1}, . . ., b_{n}$ that is divisible by every common divisor.

Proof
Use induction over $n$ , using the same argument for $n = 2$ case.
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Extending GCD for known multiples

m

For any $m \in Z^{+}$ :
$gcd (m a, m b) = m gcd (a, b)$

Proof
By properties of the GCD, we have:
$\begin{aligned} gcd (m a, m b) & = least positive value of m a x + m b y \\ = m \cdot least positive value of a x + b y \\ = m \cdot gcd (a, b) \end{aligned}$
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GCD with internal divisors

d

If $d | a$ and $d | b$ and $d > 0$ then:
$gcd (\frac{a}{d}, \frac{b}{d}) = \frac{1}{d} gcd (a, b)$
If $gcd (a, b) = g$ then:
$gcd (\frac{a}{g}, \frac{b}{g}) = 1$

Proof
(2) comes from (1) when $d = gcd (a, b)$ . Just prove (1).

(1): Use extension of GCD w/ $m$ by replacing the following: $m \leftrightarrow d$ , $a \leftrightarrow \frac{a}{d}$ , and $b \leftrightarrow \frac{b}{d}$ :
$gcd (d \cdot \frac{a}{d}, d \cdot \frac{b}{d}) = d gcd (\frac{a}{d}, \frac{b}{d}) = gcd (a, b)$
Gives the result we have above.
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Multiplicative Property of Coprimality

If $gcd (a, m) = gcd (b, m) = 1$ then $gcd (a b, m) = 1$ .

Namely, if $a, m$ are coprime and $b, m$ are coprime, then $a b$ is coprime with $m$ .

Proof
Suppose $gcd (a, m) = gcd (b, m) = 1$ . Using the Linear GCD properties then $\exists x_{0}, y_{0}, x_{1}, y_{1}$ such that:
$1 = a x_{0} + m y_{0} = b x_{1} + m y_{1}$
Rearranging:
$a x_{0} + b (- x_{1}) = m (y_{1} - y_{0})$
Notice that (starting from some new equation, using the fact that $a x_{0} = 1 - m y_{0}$ and $b x_{1} = 1 - m y_{1}$ ):
$(a x_{0}) (b x_{1}) = (1 - m y_{0}) (1 - m y_{1}) = 1 - m (y_{0} + y_{1} - m y_{0} y_{1})$
Choose $X_{0} = x_{0} x_{1}$ and $Y_{0} = y_{0} + y_{1} - m y_{0} y_{1}$ . Then:
$a b X_{0} = 1 - m Y_{0} \Rightarrow 1 = a b X_{0} + m Y_{0}$
Therefore, then using Property 3 implies that any common divisor of $a b$ and $m$ is a divisor of 1. Therefore, then $gcd (a b, m) = 1$ .
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Example

If we have $a = 10$ and $b = 21$ and $m = 11$ then we fill all GCD requirements. Furthermore, notice that:
$gcd (a b, m) = gcd (10 \cdot 21, 11) = gcd (210, 11) = 1$
This makes sense. We're essentially saying that if the primes of $a, b$ and $m$ don't match up, then combining the primes into $a b$ doesn't change the fact that they don't share prime factors with $m$ still.

Relative Primes

We say $a, b$ are relatively prime when $gcd (a, b) = 1$ . $a_{1}, . . ., a_{n}$ are all relatively prime when $gcd (a_{1}, . . ., a_{n}) = 1$ . We say they are relatively prime in pairs in case $gcd (a_{i}, a_{j}) = 1$ for all $i, j = 1, . . ., n$ when $i \neq j$ .

GCD Equivalences

For any $x \in Z$ , the following are equivalent:

$gcd (a, b)$
$gcd (b, a)$
$gcd (a, - b)$
$gcd (a, b + a x)$

Proof
Clearly (1) $\to$ (2) by reordering our integers appropriately. Similarly, we can go from (1) $\to$ (3) by just realizing that when we have $gcd (a, b) = d = a x_{0} + b y_{0} = a x_{0} + (- b) (- y_{0}) = gcd (a, - b)$ .

Start at (1) and we'll go to (4). Say $g = gcd (a, b) = d$ . We know there $\exists x_{0}, y_{0}$ such that:
$d = a x_{0} + b y_{0}$
Then:
$d = a (x_{0} - x y_{0}) + (b + a x) y_{0}$
By just rearranging terms. It follows that the GCD of $a$ and $b + a x$ is a divisor of $d$ , namely $g | d$ . If we can prove that $d | g$ then we are done and have shown $d = \pm g$ and since both are positive then $d = g$ (see Property 5).

Since $d | a$ and $d | b$ then $d | (b + a x)$ from Property 3. We know also from GCD Property 2 that every common divisor of $a$ and $b + a x$ is a divisor of their GCD, namely it's a divisor of $g$ . Hence $d | g$ .
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Example

It's pretty easy to see the first 3 as being equivalent, but what about (4)? Say $a = 180, b = 240$ so then $g = gcd (a, b) = 60$ . Then:
$gcd (180, 240 + 180 x) = 60$
Is pretty clear. That's because we could take a factor of 60 from the inside (see Mult. Property) to get:
$60 = 60 gcd (3, 4 + 3 x)$
Where the GCD portion of course becomes 1 since the $3 x$ part will make sure that the $4$ which is $1 \mod 3$ will never have that remainder be 0.

GCD to Divisibility Argument

If $c | a b$ and $gcd (b, c) = 1$ then $c | a$ .

Proof
Using Extension with $m = a$ via:
$gcd (a b, a c) = a gcd (b, c) = a$
Since we suppose that $c | a b$ and obviously $c | a c$ . Then $c | a$ because of Property 2 in conjunction with the equation above ( $c$ is a divisor of both $a b$ and $a c$ and since $gcd (a b, a c) = a$ then $c | a$ as a result via the property)
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Solving Linear Diophantine Equations

Given two integers $b, c$ , how can the GCD $g$ be found? This is hard especially for large $b, c$ , and canning the range of $bx_{0} + cy_{0}s would take forever (literally).

But we can use GCD Equivalences to help us find the GCD, as well as our values of $x_{0}, y_{0}$ (for solutions to our linear equations).

Consider $b = 963, c = 657$ . If we divide $c$ into $b$ we get $q = 1$ and remainder $r = 306$ . Thus:
$b = c q + r \Leftrightarrow r = b - c q$
In particular;
$306 = 963 - 1 \cdot 657$
Now using our GCD Equivalences, we know:
$gcd (b, c) = gcd (b - c q, c)$
By using $a = c$ and $x = - q$ in the 4th equivalence in the list (and the first $\leftrightarrow$ second of course). Thus:
$gcd (963, 657) = gcd (963 - 1 \cdot 657, 657) = gcd (306, 657)$
We've made our integers smaller! The procedure can be repeated again where now $q = 2, r = 45$ :
$= gcd (306, 657 - 2 \cdot 306) = gcd (306, 45)$
Next $q = 6, r = 36$ :
$= gcd (306 - 6 \cdot 45, 45) = gcd (36, 45) = gcd (36, 9) = 9$
Thus $gcd (963, 657) = 9$ . With this result we can write the result as a linear equation:
$\begin{aligned} 306 & = 963 - 657 \\ 45 & = 657 - 2 \cdot 306 = 657 - 2 (963 - 657) \\ = 3 \cdot 657 - 2 \cdot 963 \\ 36 & = 306 - 6 \cdot 45 = (963 - 657) - 6 \cdot (3 \cdot 657 - 2 \cdot 963) \\ = 963 - 657 - 18 \cdot 657 + 12 \cdot 963 \\ = - 19 \cdot 657 + 13 \cdot 963 \\ 9 & = 45 - 1 \cdot 36 = (3 \cdot 657 - 2 \cdot 963) - 1 \cdot (- 19 \cdot 657 + 13 \cdot 963) \\ = 22 \cdot 657 - 15 \cdot 963 \end{aligned}$
So for the solution to $g = gcd (a, b) = a x_{0} + b y_{0}$ in this case, then $g = 9$ with $x_{0} = - 15$ and $y_{0} = 22$ .

Note

The values of $x_{0}, y_{0}$ are not unique. In fact in this case $x_{0} = - 15 + 657 k$ and $y_{0} = 22 - 963 k$ are all valid solutions ( $k \in Z$ ).

Let's generalize the process:

Euclidean Algorithm

Given integers $b, c > 0$ we make a repeated application of the division algorithm:
$\begin{aligned} b = c q_{1} + r_{1} & 0 < r_{1} < c \\ c = r_{1} q_{2} + r_{2} & 0 < r_{2} < r_{1} \\ r_{1} = r_{2} q_{3} + r_{3} & 0 < r_{3} < r_{2} \\ ⋮ \\ r_{j - 2} = r_{j - 1} q_{j} + r_{j} & 0 < r_{j} < r_{j - 1} \\ r_{j - 1} = r_{j} q_{j + 1} \end{aligned}$
The $gcd (b, c)$ is $r_{j}$ , the last non-zero remainder in the division process. Values of $x_{0}, y_{0}$ where $gcd (b, c) = b x_{0} + c y_{0}$ are obtained by writing each $r_{i}$ as a linear combination of $b$ and $c$ and working your way up.

Proof
The chain of equations is obtained by dividing $c$ into $b$ , $r_{1}$ into $c$ , $r_{2}$ into $r_{1}$ , ..., $r_{j}$ into $r_{j - 1}$ . The process stops when the new remainder is 0. Thus in our application of 1 - Divisibility#^344c2e we have written the inequalities for the remainder without an equality sign:
$0 < r_{1} < c$
And we stop at the point where we'd use $\leq$ instead, so then clearly:
$0 < r_{j} < r_{j - 1} < \dots < r_{1} < c$
We'll use this to prove that $r_{j}$ is the GCD of $(b, c)$ . By 1 - Divisibility#^acc042, we observe that:
$gcd (b, c) = (b - c q_{1}, c) = gcd (r_{1}, c) = gcd (r_{1}, c - r_{1} q_{2}) = gcd (r_{1}, r_{2}) = \dots$
Continuing by mathematical induction we get that:
$gcd (b, c) = gcd (r_{1}, r_{2}) = \dots = gcd (r_{j - 1}, r_{j}) = gcd (r_{j}, 0) = r_{j}$
For showing $r_{j}$ as a linear combination, we argue by induction that each $r_{i}$ is a linear combination of $b, c$ . Clearly $r_{1}$ is such a linear combination, and so is $r_{2}$ , ... so each $r_{i}$ is a linear combination of $r_{i - 1}$ and $r_{i - 2}$ . By the inductive hypothesis we may suppose that these latter two numbers are linear combinations of $b, c$ so it follows that $r_{i}$ is.
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An example of using the algorithm is:

2

Question

Find $gcd (1819, 3587) = g$ then find integers $x, y$ that satisfy:
$1819 x + 3587 y = g$

Proof
For finding $gcd$ :
$\begin{aligned} 3587 & = 1819 \cdot 1 + 1768 \\ 1819 & = 1768 \cdot 1 + 51 \\ 1768 & = 51 \cdot 34 + 34 \\ 51 & = 34 \cdot 1 + 17 \\ 34 & = 17 \cdot 2 + 0 \end{aligned}$
So $gcd (1819, 3587) = 17$ . Solving where $b = 1819, c = 3587$ :
$1819 x + 3587 y = 17$
$\begin{aligned} 1768 & = c - 1 \cdot b \\ 51 & = b - 1 \cdot (c - b) \\ = 2 b - c \\ 34 & = 1768 - 34 (51) \\ = c - b - 34 (2 b - c) \\ = - 69 b + 35 c \\ 17 & = 51 - 1 (34) \\ = 2 b - c - 1 (- 69 b + 35 c) \\ = 71 b - 36 c \end{aligned}$
Therefore comparing the two:
$17 = 1819 (\underset{x_{0}}{\underset{⏟}{71}}) + 3587 (\underset{y_{0}}{\underset{⏟}{- 36}})$
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Least-common Multiple

The integers $a_{1}, . . ., a_{n}$ (all different from zero) have a common multiple $b$ if $a_{i} | b$ for $i = 1, 2, . . ., n$ (note that LCM's always exist, ex: $a_{1} a_{2} . . . a_{n}$ ). The least of the positive common multiples is called the least common multiple and is denoted:
$[a_{1}, . . ., a_{n}]$

LCM Equivalances

If $b$ is any common multiple of $a_{1}, . . ., a_{n}$ then $[a_{1}, . . ., a_{n}] | b$ . This is the same as saying that if $h$ denotes $[a_{1}, . . ., a_{n}]$ then $0, \pm h, \pm 2 h, . . .$ comprise all the common multiples of $a_{1}, . . ., a_{n}$

Proof
Let $m$ be any common multiple and divide $m$ by $h$ . Via 1 - Divisibility^78dbe0, there is a quotient $q$ and a remainder $r$ such that $m = q h + r$ where $0 \leq r < h$ . We must prove that $r = 0$ . Assume $r \neq 0$ . For each $i = 1, 2, . . ., n$ we know that $a_{i} | h$ and $a_{i} | m$ so then $a_{i} | r$ . Thus $r$ is a positive common multiple of $a_{1}, . . ., a_{n}$ contrary to the fact that $h$ is the least of all the positive common multiples.
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Properties of the LCM

If $m > 0$ then $[m a, m b] = m [a, b]$ and $[a, b] \cdot (a, b) = | a b |$ .

Proof
(1): Let $H = [m a, m b]$ and $h = [a, b]$ . Then $m h$ is a multiple of $m a$ and $m b$ so then $m h \geq H$ . Also, $H$ is a multiple of both $m a$ and $m b$ so $\frac{H}{m}$ is a multiple of $a$ and $b$ . Thus $\frac{H}{m} \geq h$ so then $m h = H$ .

(2): We'll prove just for positive integers $a, b > 0$ . Consider the special case where $gcd (a, b) = 1$ . Now $[a, b]$ is a multiple of $a$ , say $m a$ . Then $b | m a$ and $(a, b) = 1$ . Thus by 1 - Divisibility#^f558ea, then $b | m$ . Hence $b \leq m$ and $b a \leq m a$ . But $b a$ being positive common multiple of $b$ and $a$ cannot be less than the least common multiple, so $b a = m a = [a, b]$ .

In the general case when $(a, b) = g > 1$ we have $(\frac{a}{g}, \frac{b}{g}) = 1$ via internal divisors. Applying the result of the previous paragraph gives:
$[\frac{a}{g}, \frac{b}{g}] (\frac{a}{g}, \frac{b}{g}) = \frac{a}{g} \frac{b}{g}$
Multiplying by $g^{2}$ via 1 - Divisibility#^7ef913, in combination with (1) gives us that:
$[a, b] (a, b) = | a b |$
as desired.
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