Question

Prove -1x-1=1 using axioms
explain your steps

Answer 1

We can't prove that because if x = 2 for example, it is false.

Answer 2

no its not

Answer 3

-1(2)-1=-3 NOT 1

Answer 4

i think what this means is
\[-1\times -1=1\]

Answer 5

its a multiply symbol not an x

Answer 6

Well satellite, you just proved my point.

Answer 7

how'd i guess

Answer 8

??

Answer 9

-1 (multiplied by) -1 = 1
PROVE

Answer 10

mertsj u ever did axioms?

Answer 11

I've done axioms before but Satellite is much smarter than I so you should ask him.

Answer 12

ur smart too dont wary :)

Answer 13

what's axiom anyways?

Answer 14

the building blocks of math it is the 1st ever rules that created math

Answer 15

the point is that -1 is the additive inverse of 1, so you know how it behaves with addition, but not will multiplication
the link between multiplication and addition is the distributive property

Answer 16

pretty much

Answer 17

that is, we know that for every a there exists a -a such that a + (-a)=0

Answer 18

but that is only 1 rule though

Answer 19

so we can do better than proving
\[-1\times -1=1\] we can prove that
\[-1\times a=-a\]

Answer 20

and we do it via the distributive property
first we note that
\[a\times 0=a\times (0+0)=a\times 0+a\times 0\] the first equality because
\[0+0=0\] as 0 is the additive identity, and the second because of the distributive property
this means
\[a\times 0=a\times 0+a\times 0\] and this implies (by subtracting
\[a\times 0\] form both sides that
\[a\times 0=0\]

Answer 21

So maybe you could do something simple:
The reciprocal of -1 is -1 since 1/-1 = -1.
The product of reciprocals is 1 by the definition of reciprocals.
So (-1)(-1)=1 because they are reciprocals and the product of reciprocals is 1

Answer 22

What grade are you in Hawk?

Answer 23

now that we know
\[a\times 0=0\] we know that
\[0=a\times (1+(-1))=a\times 1+a\times (-1)\] implies that
\[-1\times a=-a\] by subtracting a from both sides

Answer 24

im in grade 11

Answer 25

So you'd better pay attention to Satellite. That little thing I did is more like grade 8 or 9

Answer 26

and now since
\[a\times -1=-a\] we know that
\[-1\times -1=1\] since -1 is the additive inverse of 1

Answer 27

in general if you want to show something that involves properties of addition and multiplication, you need the distributive law

Answer 28

dont say subtracting please cause that doesnt exist say additive inverse

Answer 29

ill hav to digest wat u just wrote...

Answer 30

wait you said the 0=0+0 ( i know this is obvious) but wat rule is this?

Answer 31

additive identity?

Answer 32

yes, 0 is the additive identity, so
\[a+0=a\] and therefore
\[0+0=0\]

Answer 33

yes but u wrote ax0=ax(0+0)?

Answer 34

isnt that assuming that 0+0=0 and not using any of the laws?

Answer 35

yes this is true because
\[0+0=0\] so
\[a\times 0=a\times (0+0)\]

Answer 36

ok

Answer 37

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