MHB Proofing Theorems with Axioms: Help Answer One!

[mermaid87]

i've been trying to prove these theorems using axioms (axioms of equality + field axioms) can anybody help me?

1. -(-a) = a
2. (-a)b = -(ab); (-1)b = -b
3. (-a)(-b) = ab
4. -(a + b) = -a + (-b)
5. If a is not 0, then 1 over 1/a = a
6. If a, b is not 0, then 1/ab = 1/a · 1/b
7. If a is not 0 and a · x = b, then x = b/a
8. a/1 = a
9. If a is not 0, then a/a = 1
10. If a, c is not 0, then b/a · d/c = bd/ac and b·c/a·c = b/a

an example of a proof would be:
Prove: a · 0 = 0

0 + 0 = 0 -- identity of axiom for multiplication
(0+0)a = 0 · a -- multiplication property of equality
a · 0 + a · 0 = a · 0 -- distributive axiom for multiplication over addition
a · 0 + a · 0 = a · 0 + 0 -- identity axiom for addition
a · 0 = 0 -- cancellation law for addition

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help answer even just one pls :(

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help answer even just one pls :(

 

[Opalg]

Hi mermaid and welcome to MHB!

Here's how I would do 1.

The definition of -x is that it satisfies the condition -x + x = 0. Apply that definition to x=-a:

-(-a) + (-a) = 0 -- definition of -(-a)
(-(-a) + (-a)) + a = a -- additive property of equality
-(-a) + ((-a) + a) = a -- associative property of addition
-(-a) + 0 = a -- definition of -a
-(-a) = a -- additive axiom for 0

 

[mermaid87]

Hi mermaid and welcome to MHB!

Here's how I would do 1.

The definition of -x is that it satisfies the condition -x + x = 0. Apply that definition to x=-a:

-(-a) + (-a) = 0 -- definition of -(-a)
(-(-a) + (-a)) + a = a -- additive property of equality
-(-a) + ((-a) + a) = a -- associative property of addition
-(-a) + 0 = a -- definition of -a
-(-a) = a -- additive axiom for 0

hello and thank you so much!

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if anybody could help me with the rest of the numbers, that would be wonderful :-)

 

[Opalg]

if anybody could help me with the rest of the numbers, that would be wonderful :-)

We're not a problem-answering service here! Our aim to to help you to work things out for yourself. So you need to do some of the work now.

For the first part of 2., start by writing down the definition of -a. Then multiply that equation on both sides by b and see where that leads you.

 

[mermaid87]

We're not a problem-answering service here! Our aim to to help you to work things out for yourself. So you need to do some of the work now.

For the first part of 2., start by writing down the definition of -a. Then multiply that equation on both sides by b and see where that leads you.

sorry :-( so i'll start with

(-a) + a = 0 -- definition of -a
(-a + a) b = 0 x b -- multiplication property of equality
-ab + ab = 0 x b -- distributive axiom of multiplication over addition ?

 

[Opalg]

so i'll start with

(-a) + a = 0 -- definition of -a
(-a + a) b = 0 x b -- multiplication property of equality
-ab + ab = 0 x b -- distributive axiom of multiplication over addition ?

Good start. The next step is to quote the example proof in your first post, which tells you that 0 · b = b · 0 = 0.

 

[mermaid87]

Good start. The next step is to quote the example proof in your first post, which tells you that 0 · b = b · 0 = 0.

ok so (-a + a) · b = 0 · b is going to be 0 · b = 0 · b using the definition of -a?

 

[mermaid87]

i'm stuck :-(

 

[Evgeny.Makarov]

The next step is to quote the example proof in your first post, which tells you that 0 · b = b · 0 = 0.

ok so (-a + a) · b = 0 · b is going to be 0 · b = 0 · b using the definition of -a?

Previously, you have showed that $(-a)b+ab=0\cdot b$. According to Opalg's advice, $0\cdot b=0$, so $(-a)b+ab=0$. Now, as in elementary arithmetic, add $-(ab)$ to both sides and rearrange the left-hand side using associativity and the property of additive inverse to get $(-a)b=-(ab)$.

 

[mermaid87]

Previously, you have showed that $(-a)b+ab=0\cdot b$. According to Opalg's advice, $0\cdot b=0$, so $(-a)b+ab=0$. Now, as in elementary arithmetic, add $-(ab)$ to both sides and rearrange the left-hand side using associativity and the property of additive inverse to get $(-a)b=-(ab)$.

i got it thank you!

i need help on the one with fractions tho :/

if a is not equal to 0, then 1 over 1/a = a
if a,b are not equal to 0, then 1/ab = 1/a · 1/b
if a is not equal to 0, then a/a = 1
if a,c are not equal to b/a · d/c = bd/ac and b·c/a·c = b/a

if i could get at least one of these, maybe i could figure out the rest? :/

 

[Evgeny.Makarov]

if a is not equal to 0, then 1 over 1/a = a

A primitive operation is multiplicative inverse: the inverse of $a$ is denoted by $a^{-1}$, and the defining property of this operation is $a\cdot a^{-1}=a^{-1}\cdot a=1$. Division is defined in terms of inverse: $a/b\overset{\text{def}}{=}a\cdot b^{-1}$. Using this definition, $1/a$=$1\cdot a^{-1}$, but using the axiom that $1\cdot b=b\cdot 1=b$ for all $b$ we get $1/a=a^{-1}$. These are just preliminary remarks.

So you need to show $\left(a^{-1}\right)^{-1}=a$. This is a multiplicative analog of $-(-a)=a$, which is already solved, and the proof is quite similar. Just replace additive inverse $-$ with multiplicative inverse $(\cdot)^{-1}$, additive unit $0$ with multiplicative unit $1$ and addition with multiplication.

 

[Deveno]

i got it thank you!

i need help on the one with fractions tho :/

if a is not equal to 0, then 1 over 1/a = a
if a,b are not equal to 0, then 1/ab = 1/a · 1/b
if a is not equal to 0, then a/a = 1
if a,c are not equal to b/a · d/c = bd/ac and b·c/a·c = b/a

if i could get at least one of these, maybe i could figure out the rest? :/

So what you want to prove is:

$\dfrac{1}{\left(\dfrac{1}{a}\right)} = a$.

First, ask yourself: what does $\dfrac{1}{a}$ even mean?

What I am trying to get you to see here, is that $\dfrac{1}{a}$ serves the same role in multiplication as $-a$ does in addition:

In addition we have:

$a + -a = 0$

in other words:

something (operation) anti-something = operation identity.

In addition, the "anti-something" for $a$ is $-a$. That is, it is some other number $b$ for which:

$a + b = 0$, where $0$ is the IDENTITY for addtion. What do we mean by "identity"? It leaves the "something" ($a$) unchanged when we add it:

$a + 0 = a$.

Now if we "change the operation" from adding to multiplying, we have a different "anti-something" called:

$\dfrac{1}{a}$ (or sometimes, the multiplicative INVERSE for $a$).

We also have a "different identity" (the thing that "does nothing" when we multiply by it), $1$:

$a \cdot 1 = a$.

So $\dfrac{1}{a}$ is some number $b$ such that: $a\cdot b = 1$.

For this $b$ (that is to say, for $\dfrac{1}{a}$), we want to show that $\dfrac{1}{b} = a$.

This is exactly parallel to showing that $-(-a) = a$, but using multiplication, instead of adding.

One small caveat: why do we insist that $a \neq 0$?

Well suppose $a = 0$. If we try to find some number $b$ such that $0\cdot b = 1$, we have (from your earlier work):

$0 = 0\cdot b = 1$, which is clearly nonsense (if $0 = 1$, then that makes ALL the numbers 0, which isn't very interesting, because we can easily imagine non-zero quantities of stuff).

What I would do is multiply:

$\dfrac{1}{a}$ and $\dfrac{1}{\left(\dfrac{1}{a}\right)}$ together (it may be conceptually easier to do this by letting (this is called "substitution"):

$b = \dfrac{1}{a}$ and using one of the axioms (rules) you already know.

This will give you an equation, which you can then mulitply both sides of by $a$.