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2.4: Arithmetic of divisibility

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Thinking out loud

If a teacher was able to share 6 apples among 6 boys and 8 oranges among 4 girls equally, is it possible for the teacher to share the 14 fruits with the 10 children equally?

Theorem: Divisibility theorem I (BASIC)

Let a,b,cZ such that a+b=c. If dZ+ divides any two of a,b and c, then d divides the third one.

Proof:

Proof:(by cases)

Case 1: Suppose da and db. We shall show that dc.

Since da and db, a=dm and b=dk, for some m,kZ.

Consider, c=a+b=dm+dk=d(m+k).

Since (m+k)Z, dc .

Case 2: Suppose da and dc. We shall show that db.

Since da and dc, a=dm and c=dk, for some m,kZ.

Consider, b=ca=d(km).

Since (km)Z,db .

Case 3: Suppose db and dc. We shall show that da.

Since db and dc, b=dm and c=dk, with m,kZ.

Consider,a=cb=d(km).

Since (km)Z, da.

Having examined all possible cases, given a + b = c, then if dZ+ divides any two of a,b and c, then d divides the third one.

Theorem: Divisibility theorem II (MULTIPLE)

Let a,b,cZ such that ab. Then abc.

Proof:

Let a,b,cZ such that ab.

We shall show that abc.

Consider that since a | b, b = ak, kZ.

Further consider bc = a(ck).

Since ckZ, a | bc.

Theorem: Arithmetic of Divisibility

Let a,b,c,dZ. Then

  1. if ab and ac then a(b+c).
  2. if ab and ac then a(bc).
  3. if ab and cd then (ac)(bd).
Proof:

Proof of 1:

Let a,b,c,dZ.

We shall show that if ab and ac then a(b+c).

Since a|b,b=ak,kZ and since a|c,c=am,mZ.

Consider, b+c=a(k+m).

Since k+mZ, a|(b+c).

Proof of 2:

Let a,b,c,dZ.

We shall show that if ab and ac then a(bc).

Since a|b,b=ak,kZ and since a|c,c=am,mZ.

Consider bc=a(akm).

Since akmZ, a|(bc).

Proof of 3:

Let a,b,c,dZ.

We shall show that if ab and cd then (ac)(bd).

Sincea|b,b=ak,kZ and since a|c,c=am,mZ.

Consider bd=(ak)(cm)=ac(km).

Since kmZ,(ac)|(bd).

Example 2.4.1:

Let a,b,c,dZ such that ab and cd. Is it always true that (a+c)(b+d) ?

In other words, if a teacher was able to share 6 apples among 6 boys and 8 oranges among 4 girls equally, is it possible for the teacher to share the 14 fruits with the 10 children equally?

Example 2.4.2:

Let a and b be positive integers such that 7|(a+2b+5) and 7|(b9). Prove that 7|(a+b).

Solution

Let a,b+s.t.7(a+2b+5) and 7(b9).

Consider a+2b+5=7(m),mZ.

Further consider b9=7(k),kZ.

Next consider a+2b+5(b9)=7m7k.

a+b+14=7(mk).

a+b=7(mk2),mk2Z.

Thus, 7|(a+b).

Example 2.4.3:

Let a and b be positive integers. Prove or disprove the following statements:

1. If a|b then a2|b3.

2. If a2|b3 then a|b.

Solution

1. This statement is true.

Proof:

Let a and b be positive integers such that  a|b. Then b=am,mZ+.

Consider b3=(am)3=a3m3=a2(am3).

Since am3Z+,a2|b3. 

2. This statement is false. Counterexample:

Choose a=33,b=32. Then  a2=36=b3.

Hence a2|b3 but  ab.

 

 


This page titled 2.4: Arithmetic of divisibility is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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