Proving!

ampohmeow · 2010-11-28 23:58:06

help me with this one;

1) Prove that a(-b) = -ab, for any real number a and b.

2) Explain why 0 is the only real number which is its own additive inverse, that is, show that if a = (-a), then a = 0.

thanks.

JaneFairfax · 2010-11-29 00:28:46

Show that (use the distributive law).

ampohmeow · 2010-11-29 01:21:48

JaneFairfax wrote:

Show that
(use the distributive law).

thanks a lot

ok let me try

ab + -(ab) = 0 existence of additive inverse?
ab + (-b)a = [b + (-b)] a by distributive law
ab + (-b)a = 0(a) additive inverse
ab + (-b)a = 0 zero product property?

therefore by uniqueness of inverse, -ab = (-b)a?

ahm i'm not sure with this

JaneFairfax · 2010-11-29 01:40:11

You are making a big mess there. What you need to do is start with the expression

You want to show that it is equal to 0. Using the distributive law gives

Can you carry on from here? What is b + (−b)?

ampohmeow · 2010-11-29 01:58:57

JaneFairfax wrote:

You are making a big mess there. What you need to do is start with the expression
You want to show that it is equal to 0. Using the distributive law gives
Can you carry on from here? What is b + (−b)?

if

ab + a(-b) = ab + a(-b) Reflexive Property of Equality (is my starting equation correct???)
= a[b+(-b)] Distributive Law
= a(0) Additive Inverse
= 0 Zero Product Property

then

a(-b) = -ab Addition Property of Equality

Last edited by ampohmeow (2010-11-29 01:59:37)

ampohmeow · 2010-11-29 02:39:00

Is my proof for #1 correct?

2) Explain why 0 is the only real number which is its own additive inverse, that is, show that if a = (-a), then a = 0.

give me idea of how to prove #2, thanks.

ampohmeow · 2010-11-29 22:49:00

anyone can help me on this?

Bob · 2010-11-30 00:02:08

hi

a = -a

add 'a' to each side

2a = 0

Now, do you have a theorem that allows you to conclude a = 0 from this ?

( I ask this because when, you are deducing properties of the real number system from basic axioms, you have to 'prove' everything else.)

Bob

Last edited by Bob (2010-11-30 01:27:49)

ampohmeow · 2010-11-30 00:09:04

bob bundy wrote:

hi
I had an idea then decided it was wrong so I've deleted it. Sorry!
I'll come back later when I've got more time for this.
Bob

thanks Bob, kindly check also my answer for #1.

Bob · 2010-11-30 01:31:35

hi

Our posts crossed over I think.

I've edited post #8 to show how I think you can do the second part.

ab + a(-b) = ab + a(-b) Reflexive Property of Equality (is my starting equation correct???)
= a[b+(-b)] Distributive Law
= a(0) Additive Inverse
= 0 Zero Product Property
then
a(-b) = -ab

There's a bit missing where you say 'then' .....

I would say you should add then inverse of 'ab' to both sides and use associativity as necessary to 'move' ab from the LHS to -ab on the RHS.

Bob

Last edited by Bob (2010-11-30 01:35:03)

ampohmeow · 2010-11-30 01:49:33

hi,

take a look

#1 problem

So,

ab + a(-b) = ab + a(-b) Reflexive Property of Equality
= a[b+(-b)] Distributive Law
= a(0) Additive Inverse
= 0 Zero Product Property

if

ab + a(-b) = 0
ab + a(-b) + (-ab) = 0 + (-ab) Addition Property of Equality
[ab+(-ab)] + a(-b) = 0 + (-ab) Associative Property
0 + a(-b) = 0 + (-ab) Additive Inverse

then
a(-b) = -ab Additive Identity

Last edited by ampohmeow (2010-11-30 02:05:46)

Bob · 2010-11-30 02:15:33

hi

ab + a(-b) + (-ab) = 0 + (-ab) Addition Property of Equality
[ab+(-ab)] + a(-b) = 0 + (-ab) Associative Property

Strictly, you've used commutativity too, as your (-ab) term has moved places and then been 're-associated'.

So how about

ab + a(-b) = 0
=> -ab +[ ab + a(-b) ] = -ab + 0 adding on the left of the expression

=> [ -ab + ab ] + a(-b) = -ab use of associativity

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

The people who set these questions like attention to little details like that.

I always found them hard because, mostly, you are proving things that are blindly obvious.

So you have to be clear about what is already proved and so can be used.

That's why I queried whether '2a = 0 => a = 0' is valid at this stage.

It depends whether you've got that proved already.

Bob

ampohmeow · 2010-11-30 02:17:12

hi
a = -a
add 'a' to each side
2a = 0
Now, do you have a theorem that allows you to conclude a = 0 from this ?
( I ask this because when, you are deducing properties of the real number system from basic axioms, you have to 'prove' everything else.)
Bob

How to show that if 2a = 0, then a = 0, i don't have here any theorem or axioms that will support the statement.

ampohmeow · 2010-11-30 02:22:46

Bob,

=> -ab +[ ab + a(-b) ] = -ab + 0 adding on the left of the expression

what property you used here? can also be like

[ab + a(-b)] + (-ab) = 0 + (-ab) Addition Property of Equality

are they just the same?

Bob · 2010-11-30 02:50:30

hi ampohmeow

In your version, you add a term on the right and then hop it across to become the middle term. OK if you want to quote 'commutative' as your reason.

Then you have to use associativity on a new line.

What I did was added the same term but on the left. Then it's in the right place to use only associativity to get the result.

As for 2a = 0, I'm stuck at the moment as I don't know what axioms you started with, and I expect it is too much for you to post them all. But I'm still thinking about it.

I'm painting my hall and landing at the moment which only takes a little bit of brain, so I'll devote the rest to the problem.

Bob

JaneFairfax · 2010-11-30 04:53:03

OMG!! What is going on? ampohmeows proof for #1 is correct!

ampohmeow wrote:

Is my proof for #1 correct?

Yes. You have proved that when a(−b) is added to ab, the result is 0. This shows that a(−b) is the additive inverse of ab (since additive inverses are unique). As the additive inverse of ab is −(ab) by definition, you have .

That is all.

ampohmeow wrote:

How to show that if 2a = 0, then a = 0, i don't have here any theorem or axioms that will support the statement.

Oh yes, you do. From the order axioms of the integers, you should be able to get a theorem that the integers form an integral domain in other words, it has no nonzero divisors of 0. Hence, from 2a = 0, you can conclude that a = 0.

Last edited by JaneFairfax (2011-01-24 07:44:14)

bobbym · 2010-11-30 11:48:32

Hi Jane;

Im sorry I didnt come to help earlier, and as a result you got bombarded with unnecessary material from Bob.
What exactly do you know about abstract algebra?
You dont need to listen to Bob and go dancing round the mulberry bush.

Everyone has feelings just like you and I. Every person wants to be treated with dignity and respect. Overposting happens all the time in a forum. It is up to the OP to choose which answer they best understand or like. It does not really matter who knows more about Abstract Algebra, you both know more than the OP.

raow · 2010-12-05 03:26:03

ampohmeow wrote:

hi
a = -a
add 'a' to each side
2a = 0
Now, do you have a theorem that allows you to conclude a = 0 from this ?
( I ask this because when, you are deducing properties of the real number system from basic axioms, you have to 'prove' everything else.)
Bob
How to show that if 2a = 0, then a = 0, i don't have here any theorem or axioms that will support the statement.

2a is like a+a. There is an axiom that can help you here.

raow · 2010-12-05 03:53:48

I am sorry, I read the problem as 2a=a.

From 2a=0 we can write 2*a=2*0, and then from cancellation (assuming you've problem that) a=0.

rwao · 2010-12-05 03:56:46

Mind, the answer depends on what system of axioms we're using. 2a=0 is not necessarily true in a ring, for example.

JaneFairfax · 2010-12-06 08:13:36

raow wrote:

From 2a=0 we can write 2*a=2*0, and then from cancellation (assuming you've problem that) a=0.

No, you cannot use the cancellation law here! The integers do not form a group under multiplication, so the cancellation law does not apply. Use the fact that the integers are an integral domain.

Last edited by JaneFairfax (2011-01-24 07:43:44)

JaneFairfax · 2010-12-06 09:24:19

A word about the cancellation law. In a group, the cancellation law applies because every element has an inverse: if a, b, c are elements of a group and ab = ac, we can multiply on the left by the inverse of a to get b = c. This does not apply to elements of a ring under multiplication because multiplicative inverses do not always exist in a ring. If a, b, c are elements of a ring and ab = ac, we cannot conclude that b = c (even if a ≠ 0). Example:

In the ring of 2×2 matrices over the integers, take

In the case of the integers, it is true that ab = ac implies b = c if a ≠ 0. The reason, however, is NOT the cancellation law; the reason is because the integers form an ordered ring. The order axioms are what force the integers to be an integral domain.

I will perhaps discuss more of this in the Euler Avenue section when I have time.

Last edited by JaneFairfax (2010-12-08 03:34:39)

Math Is Fun Forum

#1 2010-11-28 23:58:06

Proving!

#2 2010-11-29 00:28:46

Re: Proving!

#3 2010-11-29 01:21:48

Re: Proving!

#4 2010-11-29 01:40:11

Re: Proving!

#5 2010-11-29 01:58:57

Re: Proving!

#6 2010-11-29 02:39:00

Re: Proving!

#7 2010-11-29 22:49:00

Re: Proving!

#8 2010-11-30 00:02:08

Re: Proving!

#9 2010-11-30 00:09:04

Re: Proving!

#10 2010-11-30 01:31:35

Re: Proving!

#11 2010-11-30 01:49:33

Re: Proving!

#12 2010-11-30 02:15:33

Re: Proving!

#13 2010-11-30 02:17:12

Re: Proving!

#14 2010-11-30 02:22:46

Re: Proving!

#15 2010-11-30 02:50:30

Re: Proving!

#16 2010-11-30 04:53:03

Re: Proving!

#17 2010-11-30 11:48:32

Re: Proving!

#18 2010-12-05 03:26:03

Re: Proving!

#19 2010-12-05 03:53:48

Re: Proving!

#20 2010-12-05 03:56:46

Re: Proving!

#21 2010-12-06 08:13:36

Re: Proving!

#22 2010-12-06 09:24:19

Re: Proving!

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