Hints for the 3.2 assignment

Rings have the distributive property, so
(a+b)(a+-b) = (a+b)(a) + (a+b)(-b) = aa+ba+a(-b)+b(-b)
Then by thm 3.5 part 2
= aa+ba-ab-bb

If it were commutative, we could simplify this way:
aa+ab-ab-bb = aa-bb

suppose there are two zero elements: 0 and 0’. What is 0+0’?

Suppose there are two zero elements: 0 and 0’.
0 is an additive identity so 0+a=a for every element a, and in particular
0+0'=0'
0' is an additive identity so a+0'=a for every element a, and in particular
0+0'=0
Hence 0'=0+0'=0

Suppose there are two 1 elements: 1 and 1’. What is 1*1’?

Name an arbitrary element in T.
What form must it have because it is in T?

Let ab be an element in T.
What is its additive inverse?
How can you show that the additive inverse is an element in T?

The additive inverse of ab is -(ab)
You can show that the additive inverse is an element in T by writing it in the form (some element of R)b

-(ab) = (-a)b by Thm 3.5 part 2, so it is in T

Name two elements in T, then show that their sum is also in T

Let ab and cb be elements in T
Show that their sum is something times b (and is therefore in T)

ab + cb = (a+c)b which is in T
Hence T is closed under addition.

Name two elements in T, then show that their product is also in T

Let ab and cb be elements in T
Show that their product is something times b (and is therefore in T)

Note that the product of ab and cb is (ab)(cb); and multiplication is associative.

Name an element of R × 0, and show its additive inverse is in R × 0.

Let (a,0) be an element of R × 0
Which means that a is an element of R.
What is its additive inverse?
How do you know?
You can show the additive inverse is in R × 0 by showing it in the form: (something in R,0)

The additive inverse of (a,0) is (-a,0) because
(a,0)+(-a,0)=(0,0)
(-a,0) is an element of R × 0 because -a is an element of R.

Name two elements of R × 0 and show that their sum is in R × 0.

Let (a,0) and (b,0) be two elements of R × 0, where a and b are elements of R.
Simplify (a,0) + (b,0) and show it is in R × 0.

(a,0) + (b,0) = (a+b,0)
and a+b is in R, so (a+b,0) is in R × 0

Solve the equation for x.

Subtract a from both sides (add the additive inverse).

x = b + -a

Plug it in for x on the left side of the equation and simplify

a + (b + -a) = a+(-a + b) = (a+ -a) + b = 0 + b = b

Suppose there is more than one solution

Suppose there are at least two solutions, and name them

Suppose there are two solutions: c and d.
Plug both c and d in for x: what can you do with those equations?

a+c = b and a+d = b
so a+c = a+d
Now what can you do?

-a + a + c = -a + a + d
so 0 + c = 0 + d
so c = d.
Therefore, there can be only one solution.

x = a^-1b

What is the definition of a multiplicative inverse?

The definition is that x is the inverse of (ab) then
(ab)x=1 and x(ab)=1
The problem told you what should be the inverse of ab
Multiply that expression by ab on both the right and the left and simplify

When you multiply by ab on the right you write: (b^-1 a^-1)(a b) = b^-1 (a^-1 a) b = b^-1 *1* b = b^-1 b = 1
To finish part a, do the same thing again except multiply with (ab) on the left: (ab) (b^-1 a^-1) = ...

What rings do you know that aren’t commutative?

Matrices aren’t commutative under multiplication.
Do you know any invertible matrices that might not be commutative?
If there are 0’s everywhere except the main diagonal, it will be commutative with other matrices, so try for a more complicated matrix than that
Can you find the inverses of matrices? If not, Khan Academy can reteach it to you.

Here’s a matrix that doesn’t commute with everything, and that has an inverse
(1 2
3 4)

Here’s another matrix that doesn’t commute with everything, and that has an inverse
(4 3
1 1)

You should have two matrices A and B that don't commute: A × B is not equal to B × A
You will need to find the inverses A^-1 and B^-1 of both of those matrices
Next, you show that A^-1 × B^-1 is not the inverse of A × B:
you show that by multiplying out A^-1 × B^-1 × A × B.
When you multiply those 4 matrices together in that order you should get something that isn't the identity matrix. That proves that A^-1 × B^-1 is not the inverse of A × B.

You are proving that something is NOT a zero divisor

Start by supposing it is a zero-divisor, and write down what that means (using the zero divisor definition)

Assume u is a zero divisor.
That means there is an element b such that ub=0 and neither u nor b is 0.
Now you need to use what you know about units

Because u is a unit, it has an inverse v such that uv=vu=1.
How can you use v?

Multiply both sides of ub=0 by v (on the left)
v(ub) = v*0
(vu)b = 0         (associative property and thm 3.5#1)
1*b=0
b=0
This contradicts the property that b is not 0
Hence the assumption is false and u is not a 0-divisor.

Start with the first equation and do algebraic stuff to it

Move everything to the same side of the equals sign in the equation ab + ac = 0

ab + -ac = ac + -ac
ab + -ac = 0
Now write the left side as a* something

ab + a(-c) = 0         Thm 3.5#2
a(b + -c) = 0
a (b-c) = 0
What was that about zero divisors?

We know a isn’t a zero divisor. a is also not 0.
a ≠ 0, so if b-c ≠ 0 then a is a zero divisor because a(b-c)= 0
This contradicts the given property of a not being a zero divisor.
What do I know is true because of this contradiction?

That means the assumption “if b-c is not zero” is false
So, b-c=0.
Do more algebra

b + -c = 0
b + -c + c = 0 + c
b + 0 = c
b = c.