Chapter 1.8, Problem 17E

(a)

To determine

To prove: a divides b if and only if m = b .

Explanation of Solution

Given information: Let a , b and c be natural numbers, $\gcd \left(a,b\right)=d\text{ and lcm}\left(a,b\right)=m$ .

Proof:

  $(\Rightarrow )$ Suppose that a divides b . then

1. b is a common multiple of a and b , and
2. If n is a positive common multiple of b and a , then $b\le n,$ because b is the least multiple of b and so it is the least common multiple of a and b .

Hence, $\text{lcm}\left(a,b\right)=b$

  $(\Leftarrow )$ Suppose that m = b . then by definition of lcm, a divides b and b divides b .

Hence, a divides b

(b)

To determine

To prove: $m\le ab$ .

Explanation of Solution

Given information: Let a , b and c be natural numbers, $\gcd \left(a,b\right)=d\text{ and lcm}\left(a,b\right)=m$ .

Proof:

Since a divides ab and b divides ab , we see that ab is a common multiple of a and b .

Then by definition of lcm, $m\le ab$

(c)

To determine

To prove: if d = 1, then $m=ab$ .

Explanation of Solution

Given information: Let a , b and c be natural numbers, $\gcd \left(a,b\right)=d\text{ and lcm}\left(a,b\right)=m$ .

Proof:

Suppose that d = 1.

Since b divides m , $\exists k\in \mathbb{Z}$ such that $m=kb$ and so by part (b) $m=kb\le ab$ .

Then $k\le a$ .

Now, a divides $m=kb$ and d = 1 and so by Euclid’s Lemma a divides k .

Hence, $a\le k$ . But $k\le a$ and so k = a .

Therefore, $m=kb=ab$ .

(d)

To determine

To prove: if c divides a and c divides b , then $\text{lcm}\left(\frac{a}{c},\frac{b}{c}\right)=\frac{m}{c}$ .

Explanation of Solution

Given information: c divides a and c divides b .

Proof:

If c divides a and c divides b , then $\frac{a}{c}\text{ and }\frac{b}{c}$ are integers.

Since a and b divides m and c divides both a and b , we see that c divides m and so $\frac{m}{c}$ is also an integer.

Now, since a and b divide m , we see that $\frac{a}{c}\text{ and }\frac{b}{c}$ divide $\frac{m}{c}$ .

Then $\frac{m}{c}$ is a common multiple of $\frac{a}{c}\text{ and }\frac{b}{c}$ .

To show that $\frac{m}{c}$ is the least common multiple of $\frac{a}{c}\text{ and }\frac{b}{c}$ , suppose that n is another common multiple of $\frac{a}{c}\text{ and }\frac{b}{c}$ .

That is $\frac{a}{c}$ divides n and $\frac{b}{c}$ divides n . then a divides $nc$ and b divides $nc$ is a common multiple of a and b . but $\text{lcm}\left(a,b\right)=m$ and so $m\le nc$ .

Hence, $\frac{m}{c}\le n$ . That is $\frac{m}{c}$ is the least common multiple of $\frac{a}{c}\text{ and }\frac{b}{c}$ .

Therefore, $\text{lcm}\left(\frac{a}{c},\frac{b}{c}\right)=\frac{m}{c}$ .

(e)

To determine

To prove: for every natural number n , $\text{lcm}\left(an,bn\right)=mn$ .

Explanation of Solution

Given information: Let a , b and c be natural numbers, $\gcd \left(a,b\right)=d\text{ and lcm}\left(a,b\right)=m$ .

Proof:

Suppose that n is a natural number.

Since $\text{lcm}\left(a,b\right)=m,$ we see that a divides m and b divides m . then $\left(an\right)\text{ divides }\left(mn\right)\text{ and }\left(bn\right)\text{ divides }\left(mn\right)$ .

That is, $\left(mn\right)$ is a common multiple of $\left(an\right)\text{ and }\left(bn\right)$ .

To show that $\left(mn\right)$ is the least multiple of $\left(an\right)$ and $\left(bn\right)$ .

Then $\left(an\right)$ divides c and $\left(bn\right)$ divides c .

Since $\left(an\right)$ divides c , $\exists k\in \mathbb{Z}$ such that $c=\left(ak\right)n$ and so n divides c because n is an integer.

Now, since $\left(an\right)$ divides c and $\left(bn\right)$ divides c , we see that a divides $\frac{c}{n}$ and b divides $\frac{c}{n}$ .

That is $\frac{c}{n}$ is a common multiple of a and b . but $\text{lcm}\left(a,b\right)=m$ and so $m\le \frac{c}{n}$ .

Hence, $mn\le c$ .

Therefore, $\text{lcm}\left(an,bn\right)=mn$ .

(f)

To determine

To prove: $\text{gcd}\left(a,b\right)\cdot \text{lcm}\left(a,b\right)=ab$ .

Explanation of Solution

Given information: Let a , b and c be natural numbers, $\gcd \left(a,b\right)=d\text{ and lcm}\left(a,b\right)=m$ .

Proof:

Suppose that $d=\text{gcd}\left(a,b\right)$ .

Then d divide both a and b and so $\frac{a}{d}$ and $\frac{b}{d}$ are integers.

Then by part (d), $\text{lcm}\left(\frac{a}{d},\frac{b}{d}\right)=\frac{1}{d}\cdot \text{lcm}\left(a,b\right)$

But $\gcd \left(\frac{a}{d},\frac{b}{d}\right)=\frac{d}{d}=1$ .

Then by part (c) $\text{lcm}\left(\frac{a}{d},\frac{b}{d}\right)=\frac{a}{d}\cdot \frac{b}{d}$ .

Then

  $\text{lcm}\left(a,b\right)=d\cdot \text{lcm}\left(\frac{a}{d},\frac{b}{d}\right)=d\cdot \frac{a}{d}\cdot \frac{b}{d}=\frac{ab}{d}$

Hence, $d\cdot \text{lcm}\left(a,b\right)=\gcd \left(a,b\right)\cdot \text{lcm}\left(a,b\right)=ab$