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Discrete Mathematics

WEN-CHING LIEN Department of Mathematics National Cheng Kung University

2008

WEN-CHINGLIEN Discrete Mathematics

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4.3: Prime Numbers.

Definition (4.1)

If a,b∈ Zand b6=0,we say that b divides a, and we write b|a, if there is an integer n such that a=bn.When this occurs we say that b is a divisor of a, or a is multiple of b.

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Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

WEN-CHINGLIEN Discrete Mathematics

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Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

(5)

Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

WEN-CHINGLIEN Discrete Mathematics

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Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

(7)

Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

WEN-CHINGLIEN Discrete Mathematics

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Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

(9)

Theorem 4.3:

For all a,b,c ∈ Z+ a) 1|a and a|0

b) [(a|b)∧ (b|a)]ab.

c) [(a|b)∧ (b|c)]a|c d) a|ba|bx for all x∈ Z.

e) If x =y +z,for some x,y,z ∈ Z,and a divides two of the three integers x,y an z, then a divides the remaining integer.

f) [(a|b)∧ (a|c)]a|(bx+cy),for all x,y ∈ Z.(The

expression bx+cy is called a linear combination of b, c.) g) For 1≤in,let ci ∈ Z.If a divides each ci,then

a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤in.

WEN-CHINGLIEN Discrete Mathematics

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

WEN-CHINGLIEN Discrete Mathematics

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

WEN-CHINGLIEN Discrete Mathematics

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

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Lemma (4.1)

If n∈ Z+ and n id composite, then there is a prime p such that p|n.

Proof.

If not, let S be the set of all composite integers that have no prime divisor(s).

If S6= ∅,then by the Well-Ordering Principle, S has a least element m.

But if m is composite, then m=m1m2,where m1,m2∈ Z+, with 1≤m1m and 1m2m.

Since m1∈/ S,m1is prime or divisible by a prime — so, there exists a prime p such that p|m1.

Sincem=m1m2,it now follows from part(d) of Theorem 4.3 that p|m,and so S=∅.

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

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Theorem (4.4)

There are infinitely many primes.

Proof.

If not, let p1,p2, . . . ,pk be the infinite list of all primes, and letB =p1p2· · ·pk+1.

Since B >pi for all 1≤ik,B cannot be a prime. Hence B is composite.

So by lemma 4.1 there is a prime pj,where 1≤jk and pj|B.

Since pj|B and pj|p1p2· · ·pk,by Theorem 4.3(e) it follows that pj|1.

This contradiction arises from the assumption that there are only finitely many primes; the result follows.

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

WEN-CHINGLIEN Discrete Mathematics

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

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Theorem (4.5 The Division Algorithm)

If a,b∈ Z,with b>0,then there exist unique q,r ∈ Zwith a=qr +b,0≤r <b

Proof.

Let S ={atb|t∈ Z,atb>0}.If a>0 and t >0,then aS and S 6= ∅.

For a≤0,let t =a−1.Then

atb=a− (a−1)b =a(1b) +b,with(1−b)≤0, because b≥1.

So atb>0 and S 6= ∅.Hence, for any a∈ Z,S is a nonempty subset ofZ+.

By the Well-Ordering Principle, S has a least element r, where 0<r =aqb,for some q∈ Z

If r =b,then a= (q+1)b and b|a,contradicting ba.

continued...

WEN-CHINGLIEN Discrete Mathematics

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4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

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4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

WEN-CHINGLIEN Discrete Mathematics

(30)

4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

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4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

WEN-CHINGLIEN Discrete Mathematics

(32)

4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

(33)

4.5 continued.

If r >b,then r =b+c,for some c∈ Z+,and

aqb=r =b+cc=a− (q+1)b∈S,contradicting r being the least element of S. Hence, r <b.

This now establishes a quotient q and remainder r, where 0≤r <b,for the theorem. But are there other q’s and r’s that also work?

If so, let q1,q2,r1,r2∈ Zwith a=q1b+r1,for 0≤r1<b, and a=q2b+r2,for 0≤r2 <b.

Then q1b+r1 =q2b+r2b|q1q2| = |r2r1| <b, because 0≤r1,r2<b

If q16=q2,we have the contradiction b|q1q2| <b.

Hence,q1=q2,r1=r2,and the quotient and remainder are unique.

WEN-CHINGLIEN Discrete Mathematics

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Example (4.27)

Write 6137 in the octal system(base 8).Here we seek nonnegative integers r0,r1,· · ·rk,with 0<rk <8,such that 6137= (rk· · ·r2r1r0)8

6137=r0+r1·8+r2·82+· · · +rk·8k

r0=1,r1=7,r2=7,r3=3,r4=1,ri =0 for i ≥5 6137= (13771)8

(35)

Example (4.27)

Write 6137 in the octal system(base 8).Here we seek nonnegative integers r0,r1,· · ·rk,with 0<rk <8,such that 6137= (rk· · ·r2r1r0)8

6137=r0+r1·8+r2·82+· · · +rk·8k

r0=1,r1=7,r2=7,r3=3,r4=1,ri =0 for i ≥5 6137= (13771)8

WEN-CHINGLIEN Discrete Mathematics

(36)

Example (4.27)

Write 6137 in the octal system(base 8).Here we seek nonnegative integers r0,r1,· · ·rk,with 0<rk <8,such that 6137= (rk· · ·r2r1r0)8

6137=r0+r1·8+r2·82+· · · +rk·8k

r0=1,r1=7,r2=7,r3=3,r4=1,ri =0 for i ≥5 6137= (13771)8

(37)

Example (4.27)

Write 6137 in the octal system(base 8).Here we seek nonnegative integers r0,r1,· · ·rk,with 0<rk <8,such that 6137= (rk· · ·r2r1r0)8

6137=r0+r1·8+r2·82+· · · +rk·8k

r0=1,r1=7,r2=7,r3=3,r4=1,ri =0 for i ≥5 6137= (13771)8

WEN-CHINGLIEN Discrete Mathematics

(38)

Example (4.28 The Binary Number System) Base 10 Base2 Base10 Base2

0 0000 8 1000

1 0001 9 1001

2 0010 10 1010

3 0011 11 1011

4 0100 12 1100

5 0101 13 1101

6 0110 14 1110

7 0111 15 1111

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Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

WEN-CHINGLIEN Discrete Mathematics

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Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

(41)

Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

WEN-CHINGLIEN Discrete Mathematics

(42)

Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

(43)

Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

WEN-CHINGLIEN Discrete Mathematics

(44)

Example (4.31)

If n∈ Z+ and n is composite then there exists a prime p such that p|n and p≤√

n.

Proof.

Since n is composite, we can write n=n1n2,where 1≤n1n and 1n2n

We claim that one of the integers n1,n2must be less than or equal to√

n If not, then n1>√

n and n2>√

n give us the contradiction n=n1n2> (√

n)(

n) =n.

Without loss of generality ,we shall assume that n1<√ n.

If n1is prime, the result follows.

If n1is not prime, then by lemma 4.1 there exists a prime p<n1.So p|n and p≤√

n.

(45)

Thank you.

WEN-CHINGLIEN Discrete Mathematics

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