Discrete Mathematics
WEN-CHING LIEN Department of Mathematics National Cheng Kung University
2008
WEN-CHINGLIEN Discrete Mathematics
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.
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
WEN-CHINGLIEN Discrete Mathematics
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
WEN-CHINGLIEN Discrete Mathematics
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
WEN-CHINGLIEN Discrete Mathematics
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
Theorem 4.3:
For all a,b,c ∈ Z+ a) 1|a and a|0
b) [(a|b)∧ (b|a)]⇒a=±b.
c) [(a|b)∧ (b|c)]⇒a|c d) a|b⇒a|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≤i≤n,let ci ∈ Z.If a divides each ci,then
a|(c1x1+c2x2+· · · +cnxn),where xi ∈ Zfor all 1≤i≤n.
WEN-CHINGLIEN Discrete Mathematics
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≤m1≤m and 1≤m2≤m.
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=∅.
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≤m1≤m and 1≤m2≤m.
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
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≤m1≤m and 1≤m2≤m.
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=∅.
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≤m1≤m and 1≤m2≤m.
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
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≤m1≤m and 1≤m2≤m.
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=∅.
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≤m1≤m and 1≤m2≤m.
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
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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.
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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.
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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.
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≤i ≤k,B cannot be a prime. Hence B is composite.
So by lemma 4.1 there is a prime pj,where 1≤j ≤k 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
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
WEN-CHINGLIEN Discrete Mathematics
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
WEN-CHINGLIEN Discrete Mathematics
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
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 ={a−tb|t∈ Z,a−tb>0}.If a>0 and t >0,then a∈S and S 6= ∅.
For a≤0,let t =a−1.Then
a−tb=a− (a−1)b =a(1−b) +b,with(1−b)≤0, because b≥1.
So a−tb>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 =a−qb,for some q∈ Z
If r =b,then a= (q+1)b and b|a,contradicting b∤a.
continued...
WEN-CHINGLIEN Discrete Mathematics
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
WEN-CHINGLIEN Discrete Mathematics
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
WEN-CHINGLIEN Discrete Mathematics
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
4.5 continued.
If r >b,then r =b+c,for some c∈ Z+,and
a−qb=r =b+c ⇒c=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+r2⇒b|q1−q2| = |r2−r1| <b, because 0≤r1,r2<b
If q16=q2,we have the contradiction b|q1−q2| <b.
Hence,q1=q2,r1=r2,and the quotient and remainder are unique.
WEN-CHINGLIEN Discrete Mathematics
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
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
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
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
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
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≤n1≤n and 1≤n2≤n
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
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≤n1≤n and 1≤n2≤n
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.
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≤n1≤n and 1≤n2≤n
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
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≤n1≤n and 1≤n2≤n
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.
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≤n1≤n and 1≤n2≤n
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
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≤n1≤n and 1≤n2≤n
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.
Thank you.
WEN-CHINGLIEN Discrete Mathematics