Discrete Mathematics

Discrete Mathematics

WEN-CHING LIEN Department of Mathematics National Cheng Kung University

2008

4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

IF a,b∈ Z⁺and p is prime, then p|ab⇒or p|b.

proof 4.2.

If p|a, then we are finished.If not, then because p is prime, it follows that gcd(p,a) =1,and so there exist integer x,y with px +ay =1.Then b=p(bx) + (ab)y,where p|p and p|ab.So it follows from parts (d) and (e) of Theorem 4.3 that p|b.

4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

IF a,b∈ Z⁺and p is prime, then p|ab⇒or p|b.

proof 4.2.

If p|a, then we are finished.If not, then because p is prime, it follows that gcd(p,a) =1,and so there exist integer x,y with px +ay =1.Then b=p(bx) + (ab)y,where p|p and p|ab.So it follows from parts (d) and (e) of Theorem 4.3 that p|b.

4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

IF a,b∈ Z⁺and p is prime, then p|ab⇒or p|b.

proof 4.2.

If p|a, then we are finished.If not, then because p is prime, it follows that gcd(p,a) =1,and so there exist integer x,y with px +ay =1.Then b=p(bx) + (ab)y,where p|p and p|ab.So it follows from parts (d) and (e) of Theorem 4.3 that p|b.

4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

IF a,b∈ Z⁺and p is prime, then p|ab⇒or p|b.

proof 4.2.

If p|a, then we are finished.If not, then because p is prime, it follows that gcd(p,a) =1,and so there exist integer x,y with px +ay =1.Then b=p(bx) + (ab)y,where p|p and p|ab.So it follows from parts (d) and (e) of Theorem 4.3 that p|b.

4.5: The Fundamental Theorem of Arithmetic.

Lemma (4.2)

IF a,b∈ Z⁺and p is prime, then p|ab⇒or p|b.

proof 4.2.

If p|a, then we are finished.If not, then because p is prime, it follows that gcd(p,a) =1,and so there exist integer x,y with px +ay =1.Then b=p(bx) + (ab)y,where p|p and p|ab.So it follows from parts (d) and (e) of Theorem 4.3 that p|b.

Lemma (4.3)

Let a_i ∈ Z⁺,for all 1≤i ≤n.if p is prime and p|a₁a₂· · ·a_n, then p|ai for some 1≤i≤n.

Theorem (4.11)

Every integer n >1 can be written as a product of primes uniquely, up to the order of the primes.(Here a single prime is considered a product of one factor)

Lemma (4.3)

Let a_i ∈ Z⁺,for all 1≤i ≤n.if p is prime and p|a₁a₂· · ·a_n, then p|ai for some 1≤i≤n.

Theorem (4.11)

Every integer n >1 can be written as a product of primes uniquely, up to the order of the primes.(Here a single prime is considered a product of one factor)

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

If the first part is not true, let m>1 be the smallest integer not expressible as a product of primes.

Since m is not a prime, we are able to write m=m₁m₂,where 1<m₁<m,1<m₂<m.

But then m₁,m₂can be written as products of primes, because they are less than m. Consequently, with m=m₁m₂we can obtain a prime factorization of m.

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

If the first part is not true, let m>1 be the smallest integer not expressible as a product of primes.

Since m is not a prime, we are able to write m=m₁m₂,where 1<m₁<m,1<m₂<m.

But then m₁,m₂can be written as products of primes, because they are less than m. Consequently, with m=m₁m₂we can obtain a prime factorization of m.

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

If the first part is not true, let m>1 be the smallest integer not expressible as a product of primes.

Since m is not a prime, we are able to write m=m₁m₂,where 1<m₁<m,1<m₂<m.

But then m₁,m₂can be written as products of primes, because they are less than m. Consequently, with m=m₁m₂we can obtain a prime factorization of m.

proof 4.11 existence.

The proof consists of two parts: The first part covers the existence of a prime factorization, and the second part deals with its uniqueness.

If the first part is not true, let m>1 be the smallest integer not expressible as a product of primes.

Since m is not a prime, we are able to write m=m₁m₂,where 1<m₁<m,1<m₂<m.

But then m₁,m₂can be written as products of primes, because they are less than m. Consequently, with m=m₁m₂we can obtain a prime factorization of m.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and assuming uniqueness of representation for 3,4,5, . . . ,n−1,we suppose that n =p₁^s1p₂^s2· · ·p_k^sk =q₁^t1q₂^t2· · ·q_r^tr,where each p_i,1≤i ≤k,and each q_j,1≤j≤r,is a prime.Also

p1<p2< · · · <p_k and q1<q2< · · · <qr,and s_i ≥0 for all 1≤i ≤k,t_i >0 for all 1≤j≤r.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and assuming uniqueness of representation for 3,4,5, . . . ,n−1,we suppose that n =p₁^s1p₂^s2· · ·p_k^sk =q₁^t1q₂^t2· · ·q_r^tr,where each p_i,1≤i ≤k,and each q_j,1≤j≤r,is a prime.Also

p1<p2< · · · <p_k and q1<q2< · · · <qr,and s_i ≥0 for all 1≤i ≤k,t_i >0 for all 1≤j≤r.

proof 4.11 uniqueness.

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.

For the integer 2, we have a unique prime factorization, and assuming uniqueness of representation for 3,4,5, . . . ,n−1,we suppose that n =p₁^s1p₂^s2· · ·p_k^sk =q₁^t1q₂^t2· · ·q_r^tr,where each p_i,1≤i ≤k,and each q_j,1≤j≤r,is a prime.Also

p1<p2< · · · <p_k and q1<q2< · · · <qr,and s_i ≥0 for all 1≤i ≤k,t_i >0 for all 1≤j≤r.

4.11 continued.

Since p₁|n,we have p₁|q₁^t1q₂^t2· · ·qrtr.By lemma 4.3, p₁|q_j for some 1≤j ≤r.

Because p1and q_j are primes, we have p1=q_j.

In fact j=1, for otherwise q₁|n⇒q₁=p_efor some 1<e≤l and p₁<p_e=q₁<q_j =p₁.With p₁=q₁, we find that

n₁=n/p₁=p₁^s1−1p₂^s2· · ·p_k^sk =q₁^t1−1q₂^t2· · ·qrtr. Since n₁<n,by the induction hypothesis it follows that k =r,p_i =q_i for 1≤i ≤k,s₁−1=t₁−1 so (s₁=t₁), and s_i =t_i for 2≤i≤k.

Hence the prime factorization of n is unique.

4.11 continued.

Since p₁|n,we have p₁|q₁^t1q₂^t2· · ·qrtr.By lemma 4.3, p₁|q_j for some 1≤j ≤r.

Because p1and q_j are primes, we have p1=q_j.

In fact j=1, for otherwise q₁|n⇒q₁=p_efor some 1<e≤l and p₁<p_e=q₁<q_j =p₁.With p₁=q₁, we find that

n₁=n/p₁=p₁^s1−1p₂^s2· · ·p_k^sk =q₁^t1−1q₂^t2· · ·qrtr. Since n₁<n,by the induction hypothesis it follows that k =r,p_i =q_i for 1≤i ≤k,s₁−1=t₁−1 so (s₁=t₁), and s_i =t_i for 2≤i≤k.

Hence the prime factorization of n is unique.

4.11 continued.

Since p₁|n,we have p₁|q₁^t1q₂^t2· · ·qrtr.By lemma 4.3, p₁|q_j for some 1≤j ≤r.

Because p1and q_j are primes, we have p1=q_j.

In fact j=1, for otherwise q₁|n⇒q₁=p_efor some 1<e≤l and p₁<p_e=q₁<q_j =p₁.With p₁=q₁, we find that

n₁=n/p₁=p₁^s1−1p₂^s2· · ·p_k^sk =q₁^t1−1q₂^t2· · ·qrtr. Since n₁<n,by the induction hypothesis it follows that k =r,p_i =q_i for 1≤i ≤k,s₁−1=t₁−1 so (s₁=t₁), and s_i =t_i for 2≤i≤k.

Hence the prime factorization of n is unique.

4.11 continued.

Since p₁|n,we have p₁|q₁^t1q₂^t2· · ·qrtr.By lemma 4.3, p₁|q_j for some 1≤j ≤r.

Because p1and q_j are primes, we have p1=q_j.

In fact j=1, for otherwise q₁|n⇒q₁=p_efor some 1<e≤l and p₁<p_e=q₁<q_j =p₁.With p₁=q₁, we find that

n₁=n/p₁=p₁^s1−1p₂^s2· · ·p_k^sk =q₁^t1−1q₂^t2· · ·qrtr. Since n₁<n,by the induction hypothesis it follows that k =r,p_i =q_i for 1≤i ≤k,s₁−1=t₁−1 so (s₁=t₁), and s_i =t_i for 2≤i≤k.

Hence the prime factorization of n is unique.

4.11 continued.

Since p₁|n,we have p₁|q₁^t1q₂^t2· · ·qrtr.By lemma 4.3, p₁|q_j for some 1≤j ≤r.

Because p1and q_j are primes, we have p1=q_j.

In fact j=1, for otherwise q₁|n⇒q₁=p_efor some 1<e≤l and p₁<p_e=q₁<q_j =p₁.With p₁=q₁, we find that

n₁=n/p₁=p₁^s1−1p₂^s2· · ·p_k^sk =q₁^t1−1q₂^t2· · ·qrtr. Since n₁<n,by the induction hypothesis it follows that k =r,p_i =q_i for 1≤i ≤k,s₁−1=t₁−1 so (s₁=t₁), and s_i =t_i for 2≤i≤k.

Hence the prime factorization of n is unique.

Example (4.46)

Can we find three consecutive positive integers whose product is a perfect square— that is, do there exist m,n∈ Z⁺with m(m+1)(m+2) =n²?

Thank you.