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1
+
=
+−→
. We have also 1
1
lim
1
==
−−→ n
n
x
x
n
x
, therefore all the points
of the sort
n
1
, where ,...3,2,1 −−−=n , are jump discontinuity points .
0=x is a hole discontinuity point since 1
1
lim
1
lim
00
=
=
+→−→
x
x
x
x xx
.
For the readers convenience we provide the graph of )(xf .
We note that 1)(,10)( −≡ xatxxfxatxf . For all the intervals
,
1
1
1
n
x
n
n , the corresponding parts of the graph are line
segments with slopes equal to n , and for all the intervals
,...3,2,1,
1
11
=
+
−1
+
=
+−→
. We have also 1
1
lim
1
==
−−→ n
n
x
x
n
x
, therefore all the points
of the sort
n
1
, where ,...3,2,1 −−−=n , are jump discontinuity points .
0=x is a hole discontinuity point since 1
1
lim
1
lim
00
=
=
+→−→
x
x
x
x xx
.
For the readers convenience we provide the graph of )(xf .
We note that 1)(,10)( −<−=>≡ xatxxfxatxf . For all the intervals
,
1
1
1
n
x
n
<<
+
where 0>n , the corresponding parts of the graph are line
segments with slopes equal to n , and for all the intervals
,...3,2,1,
1
11
=
+
−<<− n
n
x
n
the slopes of the corresponding line segments
equal )1( +− n .
Ariel University Center Ariel University Center Ariel University Center Ariel University Center
Department of Computer Science and Mathematics
03030303����9066906690669066692692692692 Fax: adom@ariel.ac.il ::::E-mail 03-9066618 Tel:
P.O. Box 3, Ariel 44837, ISRAEL
ד"בס
Department of Computer Science and Mathematics
03030303����9066692906669290666929066692: : : : פקספקספקספקס 03030303����9066906690669066618618618618: : : : טלטלטלטל
adom@ariel.ac.il
The graph of )(xf is given below :
8 Does there exist a closed 9 – sided figure such that it is possible to draw a straight line which
intersects all of the nine sides (these intersections are not allowed to be at a vertex). The following is a
diagram of a 4 – sided closed figure with a straight line intersecting all 4 sides.
Solution: No! Any such line enters the boundary enclosed by the figure at the first point of intersection,
exits the boundary enclosed by the figure at the second point of intersection, and so on. Since the line
1 4
2
3
Ariel University Center Ariel University Center Ariel University Center Ariel University Center
Department of Computer Science and Mathematics
03030303����9066906690669066692692692692 Fax: adom@ariel.ac.il ::::E-mail 03-9066618 Tel:
P.O. Box 3, Ariel 44837, ISRAEL
ד"בס
Department of Computer Science and Mathematics
03030303����9066692906669290666929066692: : : : פקספקספקספקס 03030303����9066906690669066618618618618: : : : טלטלטלטל
adom@ariel.ac.il
begins and ends outside the boundary enclosed by the figure it must be that there are an even number of
points of intersection.
Ariel University Center Ariel University Center Ariel University Center Ariel University Center
Department of Computer Science and Mathematics
03030303����9066906690669066692692692692 Fax: adom@ariel.ac.il ::::E-mail 03-9066618 Tel:
P.O. Box 3, Ariel 44837, ISRAEL
ד"בס
Department of Computer Science and Mathematics
03030303����9066692906669290666929066692: : : : פקספקספקספקס 03030303����9066906690669066618618618618: : : : טלטלטלטל
adom@ariel.ac.il
9. Prove that if we choose 1+n distinct integers from the set of integers { }n,..,, 221 , then at least
two of them are relatively prime (have no common factor).
Solution: Partition the set into n disjoint sets each set with 2 numbers as follows:
{ } { } { } { }nnn 2,124,32,12,..,2,1 −∪∪∪= L . Since these are n disjoint sets, any choice of 1+n
distinct numbers must have 2 numbers from the same set. These 2 numbers are necessarily relatively
prime.
10. Let A and B be two square matrices of the same order which satisfy BA ≠ , BAAB = and
22 BA = . Show that the matrix BA + is singular (non invertible).
Solution: ( )( )
[ ] [ ]
0
2
22
1
22 =−=−+−=−+ BABBAABABABA . Lets assume ( ) 1−+ BA
exists. We’ll multiply the equation ( )( ) 0=−+ BABA by ( ) 1−+ BA and get 0=− BA
Which is a contradiction, therefore BA + is singular.
]1 [ BAAB =.
]2[ 22 BA =.