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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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Proposition 2.1.5. Let a and b be integers, not both zero. Then any common divisor of a and
b is a divisor of gcd(a, b).

Proof. The cast of characters in this proof:

• Integers a and b such that .

• By Proposition 1.3.8 there exists a greatest common divisor of a and b. Set g = gcd(a, b).
• An integer c such that and .

• The previous line gives rise to two more characters: The integers u and v such that and . The previous line gives also more information about c: .
The quest in this proof is . Or, more specifically the quest is and an integer z
such that .

Now we start with the proof. By Theorem 2.1.3 there exist integers x and y such that This is a quite dramatic scene, and the characters u and v demand the stage: But, the associativity of multiplication yields and distributive low now gives At this point our quest is finished in a color coordinated solution

z = ux + vy.

Since also , the quest is successfully completed.

Proposition 2.1.7. Let a and b be positive integers. Then any common multiple of a and b is
a multiple of lcm(a, b).

Proof. The cast of characters in this proof:

(I) Positive integers a and b.

(II) By Proposition 1.3.9 there exists a least positive common multiple of a and b.
Set m = lcm(a, b).

(III) The previous line, that is the phrase common multiple hides two more characters: the
integers j and k such that and .

(IV) It is important to notice the following character feature of m: It is the least positive
common multiple of a and b. What this means is the following

 If an integer x is a common multiple of of a and b and x < m, then x ≤ 0.

(V) An integer c which is a common multiple of a and b.
(VI) The previous line gives rise to two more characters: The integers u and v such that and .

The quest in this proof is . Or, more specifically the quest is and an integer z
such that .

This proposition is applied to the integers c and m > 0. By Proposition 1.4.1 there exist integers
q and r such that and .

What we learn about r from the previous line is that . But, there is more action waiting
to be unfolded here. Follow the following two sequences of equalities (all the green equalities!):

r = c − mq = au − mq = au − (aj)q = a(u − jq)
r = c − mq = bv − mq = bv − (bk)q = b(v − kq).
The conclusion is: r is a common multiple of a and b . But wait, also . Now the item
(IV) in the cast of characters (in fact the character feature of m) implies that . Since
also , we conclude . Going back to the equality , we conclude that . At this point our quest is completed in a color coordinated solution

z = q.

Proposition 2.1.10. If a and b are positive integers, then ab = gcd(a, b) · lcm(a, b).
Proof. The cast of characters in this proof:

(I) Positive integers a and b.

(II) By Proposition 1.3.9 there exists a least positive common multiple of a and b.
Set m = lcm(a, b).

(III) The previous line, that is the phrase common multiple hides two more characters: the
integers j and k such that and .

(IV) It is important to notice the following character feature of m: It is the least positive
common multiple of a and b. What this means is the following

 If an integer x is a common multiple of of a and b and x > 0, then m ≤ x.

(V) By Proposition 1.3.8 there exists a greatest common divisor of a and b. Set g = gcd(a, b).

(VI) The previous line gives rise to two more characters: The integers u and v such that and . Since and , we conclude that and .
The quest in this proof is simple .
Now we start with the proof. Consider a new green integer c = guv. Clearly

c = guv = av and c = guv = bu.

Hence and . That is c is a common multiple of a and b. Moreover, . Now
the item (IV) in the cast of characters (in fact the character feature of m) implies that .
Hence . Multiplying both sides of this inequality by g > 0 we get Hence . This is in some sense one half of the quest. For the second half, we recall
Theorem 2.1.3 and conclude that there exist integers x and y such that Multiplying both sides of this equality by m > 0 we get y . Now more characters
are demanding the scene:

mg = max + mby = (bk)ax + (aj)by = ab(kx) + ab(jy) = ab(kx + jy).

Since and , I conclude . Therefore . This is the second
half of the quest. So, the quest is completed.