Pure Mathematics proof for $(-a)b$ =$-(ab)$

Question

How can I prove this without using the Multiplicative Property of zero? $(a*0=0)$

Hint :Use Peano's Axioms - http://en.wikipedia.org/wiki/Peano_axioms — Shivam Patel, Dec 14 '13 at 05:44
What is the algebraic structure we are working with here? Are these real numbers? Are they elements of a field or ring? — Spencer, Dec 14 '13 at 05:53
They are integers. And the only axioms allowed are closure, identity, associative, commutative, and distribution. — Neel, Dec 14 '13 at 05:55
The multiplicative property of zero isn't an axiom but is in fact derivable from the axioms. Are you sure that you aren't allowed to prove that first? — Spencer, Dec 14 '13 at 06:05

achille hui · Answer 1 · 2013-12-14T19:48:13.783

14

$$\begin{array}{rll} (-a)b &= (-a)b + 0 &0 \text{ is identity for addition}\\ &= (-a)b + (ab + (-ab))&-ab \text{ is additive inverse of } ab\\ &= ((-a)b+ab) + (-ab)&\text{addition is associative}\\ &= ((-a+a)b) + (-ab)&\text{multiplication is distributive}\\ &= \color{blue}{(0b)} + (-ab)&-a \text{ is additive inverse of } a\\ &= \color{blue}{(0b + 0)} + (-ab)&0 \text{ is identity for addition}\\ &= \color{blue}{(0b + (0b + (-0b)))} + (-ab)& -0b \text{ is addition inverse of } 0b\\ &= \color{blue}{((0b + 0b) + (-0b))} + (-ab)& \text{addition is associative}\\ &= \color{blue}{((0+0)b + (-0b))} + (-ab)&\text{multiplication is distributive}\\ &= \color{blue}{(0b + (-0b))} + (-ab)&0 \text{ is identity for addition}\\ &= \color{blue}{0} + (-ab)&-0b \text{ is additive inverse of } 0b\\ &= -ab &0 \text{ is identity for addition} \end{array}$$

Please note that the part in $\color{blue}{\text{blue}}$ is a sub-proof of the statement $0b = 0$.

edited Dec 14 '13 at 19:48

answered Dec 14 '13 at 13:56

achille hui

122,701

+1, this seems to be the first post that completely answers the question without assuming that 0b = 0. – DanielV Dec 14 '13 at 14:12
I love this! This is basically the proof of 0a=0 hidden within this other proof. – Sebastian Garrido Dec 14 '13 at 15:44
@SebastianGarrido that's right. I have now colored that part in blue and hopefully this make everyone easier to see. – achille hui Dec 14 '13 at 19:50
@DanielV: I did not assume $0b=0$. – DBFdalwayse Dec 14 '13 at 21:40
Alternately to show $0b=0$ from ring axioms write $b=(0+1)b=0b+b$. – tc1729 Dec 15 '13 at 05:15
Interestingly, doesn't use multiplicative assoc, any commutivity, or multiplicative identity, making this a very general proof. If you used the assumption that $f(f^{-1}(x))$ = $x$ and that addition and subtraction are inverses, I think you can get away without even using additive association. – DanielV Dec 15 '13 at 06:40
Love this. Actually surprisingly effective and definitely smarter then whatever I could have thought of – De ath Oct 09 '16 at 23:48

score 2 · Answer 2 · edited Dec 14 '13 at 06:23

2

$-a = (-1)\cdot a$, and you have LHS $= \big((-1)\cdot a \big)\cdot b = (-1)\cdot (a \cdot b)$ by associative property = RHS

edited Dec 14 '13 at 06:23

answered Dec 14 '13 at 05:55

twenty49

29

Okay, that's what I thought. I just wasn't sure if I was allowed to do that. – Neel Dec 14 '13 at 05:58
5

Can we show $-a=(-1)\cdot a$ without using $a\cdot0=0$? – anon Dec 14 '13 at 05:59

taue2pi · Answer 3 · 2013-12-14T06:14:09.057

1

We have that $a+(-a)=0$

$\implies b(a+(-a))=0\cdot b$

$\implies ba+b(-a)=0$

$\implies b(-a)=-(ba)$

edited Dec 14 '13 at 06:14

answered Dec 14 '13 at 06:07

taue2pi

1,147
3
13
21

you can use \cdot to make a $\cdot$ for multiplication between the $0$ and the $b$. – Spencer Dec 14 '13 at 06:12
Thank you for the adivice, already edited. I'm relatively knew to LaTex – taue2pi Dec 14 '13 at 06:17
1

You make the assumption that $0 \cdot b$ = $b$, which is what the poster asked to not assume. – DanielV Dec 14 '13 at 13:25

user751379 · Answer 4 · 2020-02-16T14:37:01.287

0

by Munna shigri

a(-b)=-(ab)

proof:

0=0

b+(-b)=0. {b+(-b)=0}

a{b+(-b)}=a0. {multiplying by (a )on both side}

ab+a(-b)=a0. {by distributive property}

ab+a(-b)=0. {a0=0}

{ab+a(-b)}+(-(ab)=0+(-ab). { by adding (-ab) on both side}

{a(-b)+ab}+(-ab)=(-ab)+0. {by comutative property}

a(-b)+{ab+(-ab)}=(-ab)+0. {by associative property}

a(-b)+0=(-ab)+0. {a+(-a)=0}

a(-b)=-ab. proved {a+0=a}

edited Feb 16 '20 at 14:37

answered Feb 16 '20 at 14:29

user751379

11

1

Please use mathjax formatting. – ViktorStein Feb 16 '20 at 14:36

DBFdalwayse · Answer 5 · 2013-12-15T05:12:47.927

0

In process of rewriting. Will be back soon.

edited Dec 15 '13 at 05:12

answered Dec 14 '13 at 05:45

DBFdalwayse

999

Are you sure this won't use the property $0\cdot b=0$? – anon Dec 14 '13 at 05:56
1

Ah, you use unfair tricks like actually reading the question carefully. Let me rewrite, or delete. – DBFdalwayse Dec 14 '13 at 05:57
@anon: I think I patched it up. Please see the edit. – DBFdalwayse Dec 14 '13 at 06:05
You are still assuming $0 \cdot b = 0$. – DanielV Dec 14 '13 at 13:27
Actually, I'm not assuming it; I proved it: $0=0+0$, so$(0b)=(0+0)b=0b+0b$. Then, from $0b+0b=0b$ , I showed $0*b=0$. – DBFdalwayse Dec 14 '13 at 21:39
So, here we go again: what's wrong with the post? – DBFdalwayse Dec 15 '13 at 04:51

DanielV · Answer 6 · 2013-12-14T13:26:10.033

Suppose we only assume structure properties of distribution, closure, and existance/uniqueness of additive inverse.

$$(\forall a,b,c)\, (a + b)c = ac + bc\tag {R1}$$ Choose $b = -a$. $$(\forall a,c)\, 0 \cdot c = ac + (-a)c \tag{R2}$$

Now let's consider the possibility that both are true: $$(\exists x)\, 0\cdot x \ne 0 \tag{P1}$$ $$(\forall y,z)\, (-y)z = -(yz) \tag{P2}$$

Apply (P1) to (R2): $$(\exists x \forall a)\, 0 \ne ax + (-a)x$$ Apply (P2): $$(\exists x \forall a)\, 0 \ne ax + -(ax)$$

Which clearly contradicts closure and existance of an inverse. So for this limited structure, $\lnot (P1) \lor \lnot (P2)$.

Pure Mathematics proof for $(-a)b$ =$-(ab)$

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