Proof of (-1) (-1) = 1

We all know that (-1)(-1) = 1. But do we know how to prove this simple mathematical statement that we take for granted. In this video I tried to explain the proof!

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#maths #mathematics #number #mathematical #mathematicalproofs #numbertheory #deductivereasoning #axiom #algebra #abstractalgebra #linearalgebra #realnumbers

[mathematiciansagony]

I just want to emphasize it here, as some of you already pointed it out, while going from -(-a) to (-1) (-1) a, I used an idea that I didn't mention explicitly which is that (-1) a = (-a). You can prove it by using distributive property: x(a + b) = xa + bx.There is also a thing called 'identity element', in case of addition is 0. We can write (-1) a = (0 -1) a [as 0 is the additive identity]. By distributive property we have (0 - 1) a = 0.a - 1.a = 0 - a [By the distributive property you can prove that 0.a = 0]. Now, 0 - a = -a [Because 0 is the additive identity]. Thus (-1) a = -a. Another thing was after getting a= (-1)(-1)a, it is unnecessary to divide both sides by a. Rather you can just plug in a = 1. The good thing here is that you don't have to assume that a cannot be 0.

Those of you who are mentioning ideas from abstract algebra, please don't bring it here. The whole point of me making this video is to present it in a way so that anyone with high school algebra knowledge can get a glimpse of the proof. If I wanted to use abstract algebra here then there is no point of proving just -(-a) = a. There is a very elementary proof of the generalised version (a^-1)^-1 = a in group theory! Of course I admit that if you want to prove it very rigorous manner you have no other choice other than using ideas from abstract algebra. But, as I mentioned I didn't want to assume my audience have that prerequisite. If you have already studied abstract algebra this video will not be useful for you.Thank you everyone for watching!

[shashvatshukla]

My takeaway from the video and pinned comment:

a + (-a) = 0
a = - (-a)

To negate is to multiply (-1) because
(1-1)* a = a + (-1)*a = 0

So,
a = (-1) (-1) a
1 = (-1) (-1)

[absence]

More specifically you do utilise a range of axioms from group theory to be able to change the expression in the first place. Assuming a to be an element of the real numbers, x-x=0 is not an axiom, its a rule that follows from other axioms, here due to the definition of subtraction defined in (R, +(-), *(/), ≤).

[almatapseit]

People who don't know about this channel but are into maths and science, don't know what they're actually losing. Great video brother, keep it going! 💥

[AcryllixGD]

really good video! i thought it was brilliant and could honestly help people understand why its the case. I've always used the explanation that if you face 1 direction and turn around twice that you face the same direction and thought it was lacking but this is an actual really good explanation about why thats the case. thank you!

[pedroivog.s.]

For those who have doubts about (-1)a = -a, here it goes as a complement (please someone point it out if I forgot something):
First, we need some definitions:
Say the real numbers a, b, v, u and w. We have the following axioms of addiction (+) and multiplication (*):
1) v+w = w+v (additive commutativity)
2) (v+w)+u = v+(w+u) (additive associativity)
3) There exists a number "0" such that
0 + v = v
4) For each v, there exists a real number -v (called the opposite or additive inverse) such that
v + (-v) = 0 ( you could define subtraction here with
v+(-v)=v-v=0 )

5) v*w = w*v
6) a(bv) = (ab)v
7) 1*v = v
8) a(v+w) = av+aw (distributivity)

Now we have some proofs from these axioms:
i) there can only exist one 0 (aka a number with property 3)
Supose there exists 0_1 and 0_2 such that
v + 0_1 = v, and
0_2 + v = v
That way
0_2 = 0_1 + 0_2 = 0_2 + 0_1 = 0_1
Therefore 0_1 = 0_2. In analogous way, 1 can also be proven to be a unique number with the property 7)

ii) there can only exist one (-v) for each v.
Proof:
Suppose there are x1 and x2 such that, for some specific v:
v+x1 = 0
x2 + v = 0
That way,
v+x1 = 0
x2+(v+x1) = x2+0
(x2+v)+x1 = x2+0
0+x1=x2+0
Therefore x1 = x2.

iii) a*0 = 0
Proof:
a+ 0 = a
(a+0)*a = a*a
a*a + 0*a = a*a
(a² + 0*a) + (-a²) = a² + (-a²)
a² + [0*a + (-a²)] =a² + (-a²)
a² + [(-a²) + 0*a] = 0
[a² + (-a²)] + 0*a = 0
0 + 0*a = 0
Therefore 0*a = 0 .

Now with the theorem we want, that being:
(-1)v = -v
Proof:
v+(-1)v
= v*1 + (-1)v
= v[1+ (-1)]
= v*0 = 0
So v + (-1)v = 0.
That way, since both (-1)v and -v work as opposites for v, that means (-1)v = -v.

Wow that took way longer than I expected! It's a pretty standard proof for Linear Algebra classes (we also learn these theorems apply for any vector space, that being any vector space following those axioms whose elements are called vectors, these v, w and u I wrote on the proof, while a, b and c are still real numbers inthe "general" case), but it was nice to force myself to remember those. Hope it was helpful for anyone who read this.

[derblaue]

I think that an axiom is not necessarily self explanatory but rather a definition. First we define 0 as x+0=x and call it the neutral element of addition. Then we define -x as the inverse additiv element of x. The composition of an element with it's inverse is then the neutral element of the according operator. So x + (-x) = 0. This is only self explanatory because we choose our definitions based on how we observe reality.

[GMPranav]

Much better way imo is to use the fact that minus means turning 180 degrees in the number line, and double minus means turning 180 twice in the number line which means you are again back to the same place you started - positive. Then when a someone asks what happens if you turn only 90 degrees, you can reply "welcome to the world of complex numbers".

[kameroncole]

1.-a-(-a)=0
2.a-a=-a-(-a)
3.distribute then take the inverse of (opposite of -a)and add it to both sides
4.a+a=a+a
5.2a+2a
6.2/2leaves you with a=a or 1=1

[Daniel-inmx]

This approach seems much more complicated than it has to be. -a denotes the additive inverse of a. Additive inverse is unique: if b is an additive inverse of a, then b + a = 0, add -a to both sides, so b = -a. Now, a + -a = 0, so a = -(-a) (as -(-a) is the additive inverse of -a). We just need to show that -a = (-1) a, which follows from distributivity: 0 = a (1 + -1) = a + (-1)a, applying uniqueness, -a = (-1)a. Therefore, a = -(-a) = (-1)(-a) = (-1)((-1) a) = (-1)(-1) a. Set a=1 (no need to divide, so this works for any ring).

[nnaammuuss]

@6:02 Note: we don't need to invoke the cancellation law, as you had proved the identity for _any_ a, and therefore could take a=1 in particular. Nice work in general.

[natemajor]

Conceptually I’ve always thought if you have a negative amount of negative things well that’s positive, just makes sense to me with how things work. Really nice way to write it out I enjoyed you showing all the steps and explaining things!

[christopherflorez]

For multiplicative identity ‘a’ and inverse ‘a*’, assuming you have uniqueness of the additive inverse:

a + a* = 0
aa* + a*a* = 0a*
a* + a*a* = 0
So a*a* = a

Lemma: 0x = 0
0 + 0 = 0
0x + 0x = 0x
(0x)* + 0x + 0x = (0x)* + 0x
0x = 0

[ГеоргиГеоргиев-сг]

I start at 1*x=x,
Therefore 1=x/x, (Where x=-1) (an unused deduction)
Assuming trigonometry.
Using degrees for angles:
LHS: (-1)*(-1)=cos(180)*cos(180), by multiplication formula= cos(dif)/2+cos(sum)/2= cos(0)/2+cos(360)/2=1/2+1/2=1
RHS: 1=1.
Trig for the win.

[oguzcan]

its one nice and eager explanantion bro keep up!

[anurondas]

I can be talking totally non-sense here, but I think before proving (-1)(-1)=1, you need to prove that (-1)(-1) even exists.
I cannot think of any scenario where this will be needed, but this was essentially the problem in assuming (0.999...) exists, while proving that (0.999...) = 1 .

[abhilashtp]

That was really..what to say, simply wonderful.
Keep going bro. Liked it.

[pouyaallahyari]

Great video. Your explanation was really good and simple 👍 Thank you for sharing it with us.

[rv]

0 = 0
(-1)0 = (-1)0
(-1)0 = 0
(-1)(-1+1) = 0
(-1)(-1)+(-1)1 = 0
(-1)(-1)+(-1) = 0
(-1)(-1) = +1.


Easy peasy. And true in any unital ring.

[potato]

What I go with is M x N = area if M and N have the dimensions of length and width. Area is always positive, therefore -M x -N is always positive.

(Excluding any discussion of integrals and signed areas.)