Unformatted text preview: MATH 2345 section 54/55
Homework 12
Solutions
Must be submitted in a one-page PDF file on D2L
)()+,) . Q1. Prove by induction: 1" + 2" + ⋯ + " = (
Basis step: 1" = ( / for all ≥ 1. . ,(,+,) . / which is true since 1=1. . Inductive step: We assume that 1" + 2" + ⋯ + " = (
(2+,)(2+.) . 2" + ⋯ + ( + 1)" = ( 2(2+,) . / is true and we show that 1" + . / is true: . . ( + 1)
1 + 2 + ⋯ + ( + 1) = 1 + 2 + ⋯ + + ( + 1) = 4
5 + ( + 1)"
2 . ( + 1). 4( + 1)" . ( + 1). + 4( + 1)"
=
+
=
4
4
4
( + 1). [ . + 4( + 1)] ( + 1). [ . + 4 + 4] ( + 1). ( + 2).
=
=
=
.
2
4
4
" " " , " " , " , " , ) Q2. Prove by induction: ,×. + .×" + "×; + ⋯ + )()+,) = )+, for all ≥ 1.
, , , , Basis step: ,×. = . which is true since . = . .
Inductive step: We assume that
, , , 2+, , ,×. , , , 2 + .×" + "×; + ⋯ + 2(2+,) = 2+, and we show that + "×; + ⋯ + (2+,)(2+.) = 2+. .
.×"
1
1
1
1
+
+
+ ⋯+
( + 1)( + 2)
1×2 2×3 3×4
1
1
1
1
1
=
+
+
+ ⋯+
+
1×2 2×3 3×4
( + 1) ( + 1)( + 2) 1
( + 2)
1
=
+
=
+ + 1 ( + 1)( + 2) ( + 1)( + 2) ( + 1)( + 2)
( + 1).
( + 2) + 1 . + 2 + 1
+1
=
=
=
=
.
( + 1)( + 2) ( + 1)( + 2) ( + 1)( + 2) + 2
Q3. Prove by Induction: For every integer ∈ , we have
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· + × ( + 1) = ( + 1)( + 2) .
3 ,
,×. + 7 7 7 d. R7 is transitive.
8. a. 2. * 3 b. R 5 is not reflexive because, for example, (2, 2) Basis
1×2=
c. R5 step:
is symmetric. ,(.)(")
" ¢ R5 . which is true since 2 = 2 . d. R5 is transitive.
Inductive
step: We assume that 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· + × ( + 1) = 2(2+,)(2+.)
" ()+,)()+.)()+")
10. and
S t =we{(0,
0), that
(0,3),1 (1,0),
(1,3),+(
(2, 2),
(2,3),
(3,+3),2)(3,=O)}
show
× 2 +(1,2),
2 × 3(2,O),
+ 3 ×(3,2),
4 + 4(0,2),
× 5 +···
+ 1)
× (
.
" * 3(1,2), (2,1), (3, 2), (3,O), (OO), (0,1), (1, 1), (1,3), (3,3),
11. Tt 2.
= {(0, 2), (1,0), (2,3), (3,1), (0,3), 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· +( + 1) × ( + 2) (2,2), (2,O)}
= 1because,
× 2 + 2 for
× 3example,
+3×4+
+ × ( + 1) + ( + 1) × ( + 2)
b. R 5 is not reflexive
(2, 42)×¢5R+···
5.
(
1)( +X2)for all real numbers x, x C x. By definition of C this
13. C is not reflexive: C
is +
reflexive
c. R5 is symmetric.=
( =
(this
+ 1)
+ 2)
means that for all real numbers
x, x2 +
+ x2
1. ×
But
is false. As a counterexample, take
3
2
d.
transitive.
x=0.R5 isThenx
+x2 (
=02+02=07
+ 1)( + 2)1. 3 × ( + 1) × ( + 2) ( + 1)( + 2)[k + 3] = + = . 3 (2,O),
3 (2, x2),and
3 y C x. By
is {(0,
symmetric:
is symmetric
X for
all (0,2),
real numbers
y, if(3,x3),C (3,
y then
10. SCt =
0), (0,3),C(1,0),
(1,2),
(3,2),
(1,3),
(2,3),
O)}
2
definition of C this means that for all real numbers x and y, if X2 + y = 1 then y 2 + X2 = 1.
2
2
whether
given
relation
reflexive,
of
these.
11. Q4.
Tt
{(0,is2),true
(1,0),
(2,3),the
(3,1),
(0,3),
(1,2),isof
(2,1),
(3, 2),symmetric,
(1,orall
1),none
(1,3),
(3,3),
But=Determine
this
because
by
commutativity
addition,
x(3,O),
+ y (OO),
= transitive,
y 2 (0,1),
+ X2 for
real
numbers
.
. is
the
circle
relation
on
the
set
of
real
numbers:
For
all
, ∈
ℝ, ⇔ + = 1.
x
and
y.
(2,2), (2,O)}
C is not transitive: C is transitive t# for all real numbers x, y, and z, if x C y and y C z
2
13. C
is xnot
C is reflexive
for allthat
realfornumbers
x C x.x, By
definition
then
C reflexive:
z. By definition
of C thisX means
all real x,
numbers
y and
z, if X2 of
+ yC =this
1
2
2 that for all real numbers
2
means
x,
x
+
x2
=
1.
But
this
is
false.
As
a
counterexample,
and y + Z2 = 21 then X2 + Z = 1. But this is false. As a counterexample, take x = 0, y =take
1,
x=0.
and z =Thenx
0. Then+x2
x2 +=02+02=07
y 2 = 1 because1. 02 + 12 = 1 and y2 + Z2 =1 because 12 + 02 = 1, but
2
92 is
+ symmetric:
Z2 l 1 because
+0 2 = 0 5 X1. for all real numbers x and y, if x C y then y C x. By
C
C is0 symmetric
definition of C this means that for all real numbers x and y, if X2 + y2 = 1 then y 2 + X2 = 1.
16. But
F is this
reflexive:
any integer. Since
m-nm, =x02 and
0,2 we have that 5 1 (m - m).
25
is trueSuppose
because mbyis commutativity
of addition,
+ y = y + X2 for all real numbers
Consequently, m F m by definition of F.
x and y.
F is symmetric: Suppose m and n are any integers such that mr F n. By definition of F this
C is not transitive: C is transitive t# for all real numbers x, y, and z, if x C y and y C z
means that 5 1 (m - n), and so, by definition of divisibility, m - n = 5k for some integer k.
then x C z. By definition of C this means that for all real numbers x, y and z, if X2 + y2 = 1
Now n - m = -(m - n). Hence by substitution, n - m = -(5k) = 5 (-k). It follows that
and y 2 + Z2 = 1 then X2 + Z2 = 1. But this is false. As a counterexample, take x = 0, y = 1,
51 (n - m) by definition2 of divisibility (since
integer), and thus n F m by definition
and
z = 0. Then x2 + y = 1 because 02 + 12-k= is1 an
and y2 + Z2 =1 because 12 + 02 = 1, but
of
F.
2
2
92 + Z2 l 1 because 0 +0 = 0 5 1.
F is transitive: Suppose m, n and p are any integers such that m F n and n F p. By definition
16. Q5.
Fof is
reflexive:
m(misgiven
integer.
m-nm,
=by0 definition
and 5 transitive,
0, of
wedivisibility,
haveorthat
5m-1of
(m
-= m).
F,
this meansSuppose
that 5 1the
- any
n) and
5 1(nSince
and so,symmetric,
nthese.
5k
Determine
whether
relation
is-p),
reflexive,
none
for some integerm k,F and
n-p
51 forofsome
Consequently,
m by
definition
F. integer 1. Now m-p = (m- n) + (n-p). Hence by is the relation defined on ℤ as follows: For all , ∈ ℤ, ⇔ − is odd.
substitution,
mp
=
5k
+
51
5(k
+
1).
It follows
1(m -mrp)F by
of divisibility
F is symmetric: Suppose m and n are any
integersthat
such5 that
n. definition
By definition
of F this
(since
k
+
1
is
an
integer),
and
thus
in F p by definition of F.
means that 5 1 (m - n), and so, by definition of divisibility, m - n = 5k for some integer k.
- 17.
298 Now
- reflexive:
m = -(m0- isn).
Hence -'by for
substitution,
m 0= m.
-(5k)
5 (-k).of It0 follows
that
0 is n
not
reflexive
all integers nmn,- m
By =
definition
this means
51
(n
m)
by
definition
of
divisibility
(since
-k
is
an
integer),
and
thus
n
F
m
by
definition
Solutions
for
Exercises:
Relations
that for all integers m, m - m is odd. But this is false. As a counterexample, take any integer
of
m.F.Then m - m = 0, which is even, not odd.
Fo isis transitive:
and np are
are any
any integers
integers such
such that
that m
m 0F n.
n and
F p. By of
definition
symmetric:Suppose
Supposem,mnand
By n
definition
0 this
of
F, this
means
5 1(mand
- n)so and
5 1(n -p), ofand
m- k.n Now
= 5k
means
that
m - that
n is odd,
by definition
oddso,
m by
- ndefinition
= 2k + 1 of
fordivisibility,
some integer
for
some
integer
k,
and
n-p
51
for
some
integer
1.
Now
m-p
=
(mn)
+
(n-p).
Hence
n -m =- (m -n). Hence by substitution, n -m = -(2k + 1) = 2(-k- 1) + 1. It follows thatby
substitution,
p = 5k + 51
5(k(since+ 1). Itkfollows
5 1(m -and
p) by
definition
of divisibility
n -m is odd mby definition
of odd
-1 is that
an integer),
thus
n 0 m by
definition
(since
of 0. k + 1 is an integer), and thus in F p by definition of F.
- not reflexive:
transitive:0 0is isreflexive
transitive
#> for
integers
and By
p, if
m 0 n and
0 pmeans
then
17. 0o isis not
-' for
all all
integers
mn, m,
m n,
0 m.
definition
of 0n this
of
0
this
means
that
for
all
integers
m,
n,
and
p,
if
m
-n
is
odd
and
m
0
p.
By
definition
that for all integers m, m - m is odd. But this is false. As a counterexample, take any integer
n - p is odd then m -p is odd. But this is false. As a counterexample, take m = 1, n = 0, and
m. Then m - m = 0, which is even, not odd.
1-=
= 1 is odd and n -p = 01 =- 1 is also odd, but m -p =
p = 1. Then m -n = 1is not odd. Hence m 0 n and n O p but m q p.
19. A is reflexive: A is reflexive
reflexive property of equality. X' for all real numbers x, x 1x1 2 1. But this is true by the A is symmetric: [We must show that for all real numbers x and y, if I x
But this is true by the symmetric property of equality. 1=1 y I then yI 1=1I x1- A is transitive: A is transitive -z> for all real numbers x, y, and z, if I x 1=1 y I and I y 1=1 z ...
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