Geometry: Homework 1 Solutions

This document contains solutions to Geometry Homework 1, covering topics such as commutative property, distributive property, and proofs. It includes references to Festisov & Dubnov (2006) and Universal (2018).

Running head: GEOMETRY 1
MATH 5375 Homework 1
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MATH 5375 Homework 1
Question 1
¿ prove t h at−0=0 , we first provet h at−1∗a=−a
a∗0=0
a∗( 1+ (−1 ) ) =a+ (−a ) ¿
1∗a+ (−1 )∗a=a+ (−a ) (Multiplicative identity )
Hence−1∗( a )= (−a )
T h ereafter , we provet h at a∗0=0
a∗0=a∗0+0 (Addititive identity of 0)
a∗( 0+ 0 )=a∗0
a∗0+ a∗0=a∗0 (distributivity property)
a∗0+ a∗0+ (−a )∗0=a∗0+ (−a )∗0 ¿
a∗0=0 ( By associativity∧ properties of additive inverse )
Since−1∗a=−a∧¿
a∗0=0
Hence by transitive property ,
−0=0
Question 2 (Festisov & Dubnov, 2006)
Applying distributive property of multiplication
addition .
( a+b ) ( c+ d )= ( a+ b ) c + ( a+b ) d
Using t h e knowledge t h at multiplication iscommutative∈nature , we h ave :
( a+ b ) c+ ( a+b ) d=c ( a+b ) + d (a+b)
Applying distributive property of multiplication
addition

c ( a+b ) + d ( a+b ) =ca+ cb+da+ db
But addition is associative , so we can rearrange ¿ :
ca+ cb+ da+db=ac + ( bc+ ad ) +bd
Using t h e knowledge t h at multiplication iscommutative∈nature , we h ave :
ac + ( bc +ad ) +bd=ac+ ( ad+ bc )+ bd
Lastly , we apply associativity property of addition :
ac + ( ad +bc ) +bd=ac+ad+ bc+ad
T h us
( a+b ) ( c+ d ) =ac +ad+bc + ad
Question 3 (Uiniversal, 2018)
Suppose we have a line being divided by a point at the middle.
Thus the total length of this line is given by:
a+ b
Thereafter, we determine the square of the above sum. That is ( a+ b ) 2

¿ t h e diagram we , ( a+b )2=a2+ ab+ab+ b2=a2+ 2 ab+b2
T h us ( a+b )2=a2 +2 ab+b2
Question 4
Let A=[ 3 5
4 6 ]∧B=[5 8
9 4 ]
A+B= [3 5
4 6 ]+
[5 8
9 4 ]=
[ 8 13
13 10 ]
( A+ B )2= [ 8 13
13 10 ]2
=
[ 8 13
13 10 ]∗
[ 8 13
13 10 ]
¿ [ 8∗8+13∗13 8∗13+13∗10
13∗8+10∗13 13∗13+10∗10 ]=
[233 243
234 269 ]
( A+B )2= [233 243
234 269 ]
A2= [ 3 5
4 6 ]
2
= [ 3 5
4 6 ]∗
[ 3 5
4 6 ] =
[ 3∗3+5∗4 3∗5+5∗5∗6
4∗3+ 6∗4 4∗5+6∗6 ]= [ 29 45
36 56 ]
2 AB=2∗
[3 5
4 6 ]∗
[5 8
9 4 ]=2∗
[3∗5+5∗9 3∗8+5∗4
4∗5+6∗9 4∗8+6∗4 ]=2∗
[60 44
74 56 ]=
[120 88
148 112 ]
B2= [5 8
9 4 ]2
=
[5 8
9 4 ] *[5 8
9 4 ]= [5∗5+8∗9 5∗8+8∗4
9∗5+4∗9 9∗8+9∗8+ 4∗4 ]= [97 72
81 88 ]

A2 +2 AB+ B2= [29 45
36 56 ]+ [120 88
148 112 ]+ [97 72
81 88 ]= [246 205
265 256 ]
Since [233 243
234 269 ]≠ [246 205
265 256 ],
( A+ B ) 2 ≠ A2+2 AB+ B2
Question 5
Based on the tables above, the following are the postulates and theorems that are true:
a) Multiplication in R is commutative
0∗1=1∗0=0
b) R is closed under multiplication
0∧1 are real numbers so is 0 (1∗0=0)
c) a∗0=0
1∗0=0
d) If a∗b=0 , then either a=0∨b=0
1∗0=0
e) R is closed under addition
1+0=1 …The resultant 1 is real number because 1∧0 are real numbers
f) Addition in n R is commutative
0+1=1+ 0=1
Question 6
f ( n ) be t h e statement t h at :1+2+3+ …+n=n ( n+1 )
2
First , we prove t h at f ( 1 ) istrue .
f ( 1 ) =1 ( 1+1 )
2 =2
2 =1.
T h us f ( 1 ) is true .
T h ereafter , we provet h at if t h ere an m suc h t h at f ( m ) istrue , t h en f ( m+1 ) must be true .

Hence f ( m ) will be represented by :1+ 2+ …+ m ( m+1 )
2 ( By inductive assumption )
T h e task ¿ is determine w h et h er f ( m+1 ) is true∨not .
f ( m+1 ) :1+2+…+m+ ( m+1 ) = ( m+ 1 ) ( m+1+1 )
2 = ( m+1 ) ( m+2 )
2
Let k =m+1
T h us ( m+1 ) ( m+1+1 )
2 = k ( k +1 )
2 .
T h is s h ows t h at f ( 1 ) , f ( m ) ∧f ( m+1 ) h olds .
Hence t h e prove
Question 7
Given that A is an empty set, then A cannot be a proper subset of A. Hence A
⊂ A ∩ B does
not hold for all sets.

References
Festisov, A., & Dubnov, Y. (2006). Proof in Geometry: With Mistakes in Geometric Proofs.
Dover Publications: New York.
Uiniversal. (2018). Understanding Geometry Proofs. Retrieved from Universal:
https://www.universalclass.com/articles/math/geometry/understanding-geometry-
proofs.htm