Positives and Negatives

[A Fool]

It's a Basic Question and therefore should be asked if other things are built upon it.

 

why is a negative times a negative a positive and a negative times a positive a negative and similarily for division?

 

If i missed something basic back in grade 3 please be gentle

thanks,

 

 

A Fool

[the tree]

Multiplying a number by -1, if you try to imagine it visually, reflects the number about the 0 on the number line.

As you'll know if you've done some geometry, reflecting something about the same point twice, leaves it as it is.

[cosine]

Hm, thats the first time I've heard it explained that way tree, I like it!

 

An algebraic way of looking at it could go this way:

For a positive*negative, multiplication is defined as addition of one number, summing with itself the number of times as the other number. So (-1)*a = -1 + -1 + -1... a times.

 

For neg*neg here is a proof by contradiction:

assume (-a)*(-b) =/= a*b

then (-a)*(-b) - a*b =/= 0

so (-a)*(-b) + (-a)*b =/= 0

by the distributive law (-a)(-b + b) =/= 0

(-a)(0) = 0 =/= 0 is a contradiction.

 

Therefore (-a)*(-b) = a*b

 

And division works the same way because division is just multiplication by a multiplicative inverse, which has the same sign.

[Tom Mattson]

Here's another proof, not by contradiction.

 

Let [math]a,b,c\in\mathbb{R}[/math].

 

Proposition 1

[math]a+c=b+c[/math] implies [math]a=b[/math].

 

(Actually this could be an "iff" statement, but I don't need the converse to get to what I want.)

 

Proof

[math]a+c=b+c[/math] (by hypothesis)

[math](a+c)+(-c)=(b+c)+(-c)[/math] (add [math]-c[/math] to both sides)

[math]a+(c+(-c))=b+(c+(-c))[/math] (by associativity under addition)

[math]a+0=b+0[/math] (by the additive inverse property)

[math]a=b[/math] (by the additive identity property)

 

Proposition 2

[math]0a=a0=0[/math]

 

Proof

[math]a0+0=a(0+0)[/math] (by the additive identity property)

[math]a0+0=a0+a0[/math] (by the left distributive property)

[math]0=a0[/math] (by proposition 1)

[math]0=0a[/math] (by commutativity under multiplication)

 

Proposition 3

[math](-a)b=-(ab)[/math]

 

Proof

[math]a+(-a)=0[/math] (by the additive inverse property)

[math](a+(-a))b=0b[/math] (multiply both sides by [math]b[/math])

[math]ab+(-a)b=0b[/math] (by the right distributive property)

[math]ab+(-a)b=0[/math] (by proposition 2)

[math](-a)b=(-ab)[/math] (by the additive inverse property)

 

Proposition 4

[math](-a)(-b)=ab[/math]

 

Proof

[math]a+(-a)=0[/math] (by the additive inverse property)

[math](a+(-a))(-b)=0(-b)[/math] (multiply both sides by [math](-b)[/math])

[math]a(-b)+(-a)(-b)=0(-b)[/math] (by the right distributive property)

[math]a(-b)+(-a)(-b)=0[/math] (by proposition 2)

[math]-ab+(-a)(-b)=0[/math] (by proposition 3)

[math](-a)(-b)=ab[/math] (by the additive inverse property)

 

Note: At the end of the proofs for Propositions 3 and 4, I relied on the fact that the additive structure of a field is a group, and that every element in a group has a unique inverse element.

[A Fool]

How very thouroughly enlightening thank you very much!

[cosine]

Here's another proof, not by contradiction.

 

Let [math]a,b,c\in\mathbb{R}[/math].

 

Proposition 1

[math]a+c=b+c[/math] implies [math]a=b[/math].

 

(Actually this could be an "iff" statement, but I don't need the converse to get to what I want.)

 

Proof

[math]a+c=b+c[/math] (by hypothesis)

[math](a+c)+(-c)=(b+c)+(-c)[/math] (add [math]-c[/math] to both sides)

[math]a+(c+(-c))=b+(c+(-c))[/math] (by associativity under addition)

[math]a+0=b+0[/math] (by the additive inverse property)

[math]a=b[/math] (by the additive identity property)

 

Proposition 2

[math]0a=a0=0[/math]

 

Proof

[math]a0+0=a(0+0)[/math] (by the additive identity property)

[math]a0+0=a0+a0[/math] (by the left distributive property)

[math]0=a0[/math] (by proposition 1)

[math]0=0a[/math] (by commutativity under multiplication)

 

Proposition 3

[math](-a)b=-(ab)[/math]

 

Proof

[math]a+(-a)=0[/math] (by the additive inverse property)

[math](a+(-a))b=0b[/math] (multiply both sides by [math]b[/math])

[math]ab+(-a)b=0b[/math] (by the right distributive property)

[math]ab+(-a)b=0[/math] (by proposition 2)

[math](-a)b=(-ab)[/math] (by the additive inverse property)

 

Proposition 4

[math](-a)(-b)=ab[/math]

 

Proof

[math]a+(-a)=0[/math] (by the additive inverse property)

[math](a+(-a))(-b)=0(-b)[/math] (multiply both sides by [math](-b)[/math])

[math]a(-b)+(-a)(-b)=0(-b)[/math] (by the right distributive property)

[math]a(-b)+(-a)(-b)=0[/math] (by proposition 2)

[math]-ab+(-a)(-b)=0[/math] (by proposition 3)

[math](-a)(-b)=ab[/math] (by the additive inverse property)

 

Note: At the end of the proofs for Propositions 3 and 4, I relied on the fact that the additive structure of a field is a group, and that every element in a group has a unique inverse element.

Didn't all of your proofs rely on the converse of proposition 1 at some point?

[Tom Mattson]

Yes, I supposed they do. But I immediately have the converse from the definitions of addition and of equality. If [math]a=b[/math], then I have:

 

[math]a+c=a+c[/math] (tautology)

[math]a+c=b+c[/math] (substitution)

 

So there we have the converse, and everything is now logically closed.

[bombus]

Multiplying a number by -1, if you try to imagine it visually, reflects the number about the 0 on the number line.

As you'll know if you've done some geometry, reflecting something about the same point twice, leaves it as it is.

 

Great explanation. Now I geddit!

[cosine]

Yes, I supposed they do. But I immediately have the converse from the definitions of addition and of equality. If [math]a=b[/math], then I have:

 

[math]a+c=a+c[/math] (tautology)

[math]a+c=b+c[/math] (substitution)

 

So there we have the converse, and everything is now logically closed.

Nice!

[uncool]

Tom - not quite; substitution isn't quite an axiom. However, the converse that you're stating actually itself is an axiom - the axiom of well-definedness, isn't it?

=Uncool-