Proof that minus a minus is positive

[Jim77]

Is there a proof of why - -a =+a

In other words minus minus a is positive a?

 

[Dr.Peterson]

Is there a proof of why - -a =+a

In other words minus minus a is positive a?

Yes.

The definition of "-x" is this:

The negation (additive inverse) of x is the number y such that x + y = 0. We write, -x = y.​

Therefore, the negation (additive inverse) of -a is the number y such that -a + y = 0. Since this is true for y = a, we conclude that -(-a) = a.

You can also just think of negation as flipping the number line around, so that every positive number becomes the corresponding negative number. But this also turns every negative number into the corresponding positive number: -(5) = -5; -(-5) = 5.

 

[stapel]

Is there a proof of why - -a =+a

In other words minus minus a is positive a?

There are plenty of informal ways of looking at this. For instance, "a bad thing happens to a bad person" is a good thing, so the negative ("a bad thing") applied to a negative ("a bad person") is a positive ("a good thing"). Another way would be to note that, if you remove a cold-pack from a cold thing, the thing gets warmer (using "cold" as "negative" and "warm" as "positive").

One actual proof is done "by contradiction". This means that you assume the opposite of what you're trying to prove, do completely valid mathematics to this opposite thing, and end up with garbage. Since the only suspect part was the assumption, then ending up with garbage contradicts the assumption, and thereby proves that what you'd assumed the opposite of is the correct version.

So let's do that. We are wanting to see if it can be proven that [imath]-(-a) = +a[/imath]. The minus-of-minus expression has to be positive or negative. The claim is that it's positive; let's assume the opposite. Let's assume that [imath]-(-a) = -a[/imath].

(For the proof to work, I will assume that you are familiar with the Distributive Law, which says that [imath]x(y + z) = xy + xz[/imath]. Also, I will assume that we agree that multiplying by zero gives you zero, and that [imath]x - x = 0[/imath] regardless of what [imath]x[/imath] might be.)

Then we can do the following:

[imath]\qquad 0 = 0(-1)[/imath]

[imath]\qquad \qquad = (1- 1)(-1)[/imath]

[imath]\qquad \qquad = (1 + (-1))(-1)[/imath]

[imath]\qquad \qquad = (1)(-1) + (-1)(-1)[/imath]

[imath]\qquad \qquad = -1 + (-1)[/imath]

[imath]\qquad \qquad = -2[/imath]

But we'd started with zero, and [imath]-2[/imath] is very much *not* equal to zero. We did perfectly valid math on each step, but we ended up with garbage.

The only suspect part of the working above is the assumption that [imath]-(-1) = -1[/imath]. Since that assumption led to nonsense, then the assumption must be wrong. Therefore, it must be true that [imath]-(-1) = +1[/imath]. Since [imath]-a = (-1)(a)[/imath], we can see that [imath]-(-a)[/imath] must be equal to [imath]+a[/imath].

 

[Steven G]

Here is my proof.
+a + x = 0 means that x = -(a) by definition.
That is, what ever you add to a to get 0 is defined to be (-a).

Now (-a) is some number. Whatever you add to (-a) to get zero will be called -(-a). Since (-a) + a = 0, then we have that a = -(-a)

 

[Steven G]

"a bad thing happens to a bad person" is a good thing, so the negative ("a bad thing") applied to a negative ("a bad person") is a positive ("a good thing").

"a bad thing happens to a bad person" is a good thing
I hope that you really don't mean that!

 

[stapel]

"a bad thing happens to a bad person" is a good thing
I hope that you really don't mean that!

Not literally (usually); it's more the concept. I mean, a murderer *should* go to jail, right? And giving the family some justice while also removing a murderer from polite society is surely a less-than-terrible outcome?

 

[Aion]

Is there a proof of why - -a =+a

In other words minus minus a is positive a?

I remember "learning" this in school, however, it was stated as a fact and no proof was given. Most textbooks at the pre-algebra level don't deem it necessary to prove this statement and more advanced textbooks generally just prove this as a consequence of the field axioms for real numbers. I think this is completely wrong of doing things. You should first present a theory of whole numbers, then define fractions from previous theorems about whole numbers, and then look at a collection of numbers to both left and right of 0. The fractions together with their reflected images will then form a new number system that preserves the four arithmetic operations as was previously defined for fractions. This number system we call rational numbers, where rational numbers or more specifically negative numbers, are a new and more abstract concept. And when we introduce these new numbers they have to be consistent with the original arithmetic operations among whole numbers. Clearly we need the concept of a vector to define the addition of rational numbers. Then many properties of the rational numbers become self-evident. For example, we could prove the following theorem. Given two vectors [imath]\vec{x}[/imath] and [imath]\vec{y}[/imath], where [imath]x[/imath] and [imath]y[/imath] are rational numbers, are equal if they have the same length and direction. Then we could proceed to show that the addition of vectors is commutative, namely that [imath]\vec{x}+\vec{y}=\vec{y}+\vec{x}[/imath]. We could then define subtraction for rational numbers as addition such that if [imath]x[/imath] and [imath]y[/imath] are rational numbers then [imath]x-y=x+y^*[/imath], where [imath]y^*[/imath] is the reflected image of [imath]y[/imath]. Then by definition [imath]0-y=y^*[/imath], such that [imath]-y=y^*[/imath] and we call [imath]-y[/imath] minus [imath]y[/imath] or the opposite of [imath]y[/imath] etc. I'm well aware that the reasoning necessary to conclude that [imath]-(-(a))=a[/imath] for a rational number [imath]a[/imath] would be rather long, although I feel like it would be more intuitive than to derive it from axioms of real numbers.

 

[Aion]

I remember "learning" this in school, however, it was stated as a fact and no proof was given. Most textbooks at the pre-algebra level don't deem it necessary to prove this statement and more advanced textbooks generally just prove this as a consequence of the field axioms for real numbers. I think this is completely wrong of doing things. You should first present a theory of whole numbers, then define fractions from previous theorems about whole numbers, and then look at a collection of numbers to both left and right of 0. The fractions together with their reflected images will then form a new number system that preserves the four arithmetic operations as was previously defined for fractions. This number system we call rational numbers, where rational numbers or more specifically negative numbers, are a new and more abstract concept. And when we introduce these new numbers they have to be consistent with the original arithmetic operations among whole numbers. Clearly we need the concept of a vector to define the addition of rational numbers. Then many properties of the rational numbers become self-evident. For example, we could prove the following theorem. Given two vectors [imath]\vec{x}[/imath] and [imath]\vec{y}[/imath], where [imath]x[/imath] and [imath]y[/imath] are rational numbers, are equal if they have the same length and direction. Then we could proceed to show that the addition of vectors is commutative, namely that [imath]\vec{x}+\vec{y}=\vec{y}+\vec{x}[/imath]. We could then define subtraction for rational numbers as addition such that if [imath]x[/imath] and [imath]y[/imath] are rational numbers then [imath]x-y=x+y^*[/imath], where [imath]y^*[/imath] is the reflected image of [imath]y[/imath]. Then by definition [imath]0-y=y^*[/imath], such that [imath]-y=y^*[/imath] and we call [imath]-y[/imath] minus [imath]y[/imath] or the opposite of [imath]y[/imath] etc. I'm well aware that the reasoning necessary to conclude that [imath]-(-(a))=a[/imath] for a rational number [imath]a[/imath] would be rather long, although I feel like it would be more intuitive than to derive it from axioms of real numbers.

If anyone is curious how you would prove [imath]-(-(x))=x[/imath] from the above definition, it's actually very short. From applying the same mirror reflection * operation twice as in [imath]x^{**}=x[/imath] for any [imath]x \in \mathbb{Q}[/imath], we get [imath]-(-(x))=x[/imath]. This is not my proof as I read it in a book called "Teaching School Mathematics Pre-Algebra by Hung-Hsi Wu" years ago and I highly recommend it if your interested in more details.

 

[Dr.Peterson]

If anyone is curious how you would prove [imath]-(-(x))=x[/imath] from the above definition, it's actually very short. From applying the same mirror reflection * operation twice as in [imath]x^{**}=x[/imath] for any [imath]x \in \mathbb{Q}[/imath], we get [imath]-(-(x))=x[/imath]. This is not my proof as I read it in a book called "Teaching School Mathematics Pre-Algebra by Hung-Hsi Wu" years ago and I highly recommend it if your interested in more details.

In simpler terms, I said essentially the same thing:

You can also just think of negation as flipping the number line around, so that every positive number becomes the corresponding negative number. But this also turns every negative number into the corresponding positive number: -(5) = -5; -(-5) = 5.

I presume Wu, in a pre-algebra context, would be similarly elementary.

What counts as proof depends on context.

 

[Aion]

In simpler terms, I said essentially the same thing:

I presume Wu, in a pre-algebra context, would be similarly elementary.

What counts as proof depends on context.

I suppose your right, but Wu actually does something I think is nontrivial for a beginner. He uses a novel idéa to treat rational numbers as vectors, then uses this definition to prove all the properties of rational numbers. For given that a number is a point on the number line. We look at all numbers together, i.e. numbers that could be on either side of [imath]0[/imath] and, in particular do not have to be fractions. Two numbers are said to be equidistant from [imath]0[/imath] if their distances from [imath]0[/imath] are equal. Given a number [imath]p\neq0[/imath], denote [imath]p^*[/imath] the mirror reflection of [imath]p[/imath] with respect to [imath]0[/imath], i.e. [imath]p[/imath] and [imath]p^*[/imath] are equidistant from [imath]0[/imath] and are opposite sides of [imath]0[/imath]. If [imath]p=0[/imath], let [imath]0=0^*[/imath], then for any point [imath]p[/imath], it is clear that [imath]p^{**}=p[/imath]. Which expresses the fact that reflecting a nonzero point across [imath]0[/imath] twice in succession brings it back to itself. However it still remains to be proven the relationship between addition and the mirror reflection *. Namely that [imath](x+y)^*=x^*+y^*[/imath] and [imath]x^*y^*\neq (xy)^*[/imath] for rational numbers [imath]x[/imath] and [imath]y[/imath].