quiz 3 (groups and rings)
Name: DrSarah Greenwald
Start Time: Apr 12, 2000 16:41 Time Allowed: 20 min
Number of Questions: 3

Question 1  (2 points)

For a completely correct proof, must we always follow the exact order of the statement we are trying to prove, even if it is an awkward order?

1. yes  
2. no  

Question 2  (14 points)

I've been very sloppy in my proofwriting and left out the first sentance which will tell you what I'm trying to prove, and the last sentance which tells you what I've proven, but at least I've included a list of statments with my proofs.

Match the statements on the left with the incomplete proof on the right that (when completed) proves or disproves the statement.

a,b,c in R, a not 0, ab=ac implies b=c   Let R=Z_6={0,1,2,3,4,5}. Take a=2, b=1, c=4. Notice that a,b,c in R and a not 0. Also, ab=2*1=2=2mod6=8mod6=2*4=ac, but 1 is not equal to 4, as desired.
a,b, in R, ab=0 implies a=0 or b=0   Let R=Z_6={0,1,2,3,4,5}. Take a=0, b=1. Notice that a,b in R, ab=0*1=0 and a=0, as desired.
a in R implies, a not 0, a^2 = a implies a=0 or a=1.   Let R=Z_6={0,1,2,3,4,5}. Take a=1. Notice a in R, a not 0, a^2=1^2=1, and a=1, as desired.
There exists R, there exists a,b, in R s.t. ab=0 and (a=0 or b=0)   Let R=Z_6={0,1,2,3,4,5}. Take a=2, b=3. Notice that a,b in R and ab=2*3=6=0mod6 and 2 is not 0 and 3 is not 0, as desired.
There exists R, there exists a in R s.t. a not 0, a^2=a and (a=0 or a=1)   Let R=Z_6={0,1,2,3,4,5}. Take a=4. Notice a in R, a not 0, a^2=4^2=16=4mod6 and a not 0 and a not 1, as desired.
There exists R, there exists a,b,c in R s.t. a not 0, ab=ac and b=c.   Let R=Z_6={0,1,2,3,4,5}. Take a=2, b=1, c=1. Notice a,b,c in R, a not 0, ab=2*1=2=2*1=ac and b=1=c, as desired.
a,b,c in R, a not 0, ab=ac implies b=c -->
a,b, in R, ab=0 implies a=0 or b=0 -->
a in R implies, a not 0, a^2 = a implies a=0 or a=1. -->
There exists R, there exists a,b, in R s.t. ab=0 and (a=0 or b=0) -->
There exists R, there exists a in R s.t. a not 0, a^2=a and (a=0 or a=1) -->
There exists R, there exists a,b,c in R s.t. a not 0, ab=ac and b=c. -->

Question 3  (4 points)

Notice that in the following statements on the left, the type and order of quantifiers matters very much since switching them around results in a statement that is logically very different than the original.

Match the statement to the proof that either proves or disproves it.

1. For all x in R\{0}, there exists y in R s.t. xy=1
   a. Let x be an arbitrary element of R\{0}. We will produce y in R s.t. xy=1. Take y=1/x. Notice that 1/x is in R since x is not 0 by def. Also,
xy=x(1/x)=x/x=1, as desired.
2. There exists y in R s.t. for all x in R\{0}, xy=1
   b. Let y in R be arbitrary. We will produce x in R\{0} s.t. xy is not 1.
Case 1 y not 0
Take x=2/y. Notice that x is in R\{0} since y is not equal to 0. Also, xy=2/y(y)=2(y/y)=2(1)=2. Since 2 is not equal to 1, we have shown that xy is not equal to 1.
Case 2 y=0
Take x=3. Notice that x is in R\{0} since 3 is in R\{0}. Also, xy=3(0)=0. Since 0 is not equal to 1, we have shown that xy is not equal to 1.
So, for each y in R, we have produced x in R\{0} so that xy is not equal to 1, as desired.
3. There exists x in R\{0} s.t. for all y in R, xy=1
   c. We will produce y in R s.t. for all x in R\{0}, xy is not equal to 1. Take y=0. Notice that y is in R. Let x be in R\{0}. We will show that xy is not equal to 1. Notice that xy=x(0)=0, by multiplication by zero. Since 0 is not equal to 1, we have shown that xy is not equal to 1, as desired.
4. For all y in R, there exists x in R\{0} s.t. xy=1
   d. Let x be an arbitrary element of R\{0}. We will produce y in R s.t. xy is not 1. Take y=0. Notice that y is in R. Also, xy=x(0)=0 by multiplication by 0. Since 0 is not equal to 1, we have shown xy is not equal to 1, as desired.
1 -->
2 -->
3 -->
4 -->