# What does the minus sign as a affix mean?

1. Aug 4, 2013

### Atran

Hi;

According to Wiki the minus sign is a unary operator (a negator). Calculations are done right even if the minus sign is considered to be a (-1) multiplier.

How to explain, for instance, -(a+b)=-a+-b if the sign is considered an operator?

Thanks...

2. Aug 4, 2013

### SteamKing

Staff Emeritus
The distributive property of multiplication. The leading minus sign is equivalent to multiplication by -1.

3. Aug 4, 2013

### D H

Staff Emeritus
If multiplication exists, that is.

That -(a+b) = (-a) + (-b) is true even in a group with addition (i.e., multiplication is not defined) plus an additive inverse. You don't need to invoke multiplication.

4. Aug 4, 2013

### D H

Staff Emeritus
That adds *nothing* to the discussion, pun intended.

To the OP, you can show that -(a+b) = -a + -b without invoking multiplication. All you need is that
• a+(-a)=0 for all elements in the group (definition of additive inverse),
• addition is closed (so a+b exists),
• addition is associative (a+(b+c)=(a+b)+c), and
• equality is transitive (if a=b and b=c then a=c).

5. Aug 4, 2013

### Atran

Should I think this way: if (a+b)+-(a+b) = 0, then a+b+-(a+b) = 0, therefore -(a+b)=-a+-b

I've found this book (https://www.amazon.com/books/dp/082182693X), but its level seems above mine.
Is there a book you can recommend with a chapter which proves all the arithmetical/algebraic rules?

Thanks for help.

Last edited by a moderator: May 6, 2017
6. Aug 4, 2013

### Atran

What about b-a=-(a-b)? The idea of thinking of minus as a (-1) multiplier makes it easier to work with and manipulate algebraic expressions. If the minus sign is instead a unary operator, then why so many (except Wiki) sources don't mention that?

7. Aug 4, 2013

### D H

Staff Emeritus
Google the term "additive inverse" and you'll see lots and lots of page, including wikipedia, that cover this topic.

8. Aug 5, 2013

### coolul007

unfortunately, more and more the minus sign is being interpreted as an operator. Numbers carry a sign therefore the distinction between -2 and -(2) is getting more and more blurred. recent text books as well as "smart" calculators have -2^2 = -4. and only have (-2)^2 = +4. Numbers should carry a sign, I believe that is part of the definition of additive inverse, nothing really to do with subtraction( which shouldn't exist).

9. Aug 5, 2013

### lurflurf

The confusion here is that - is overloaded. That is the same symbol is used for different purposes.
-2
might mean
the number -2
the additive inverse of the number 2
the product (-1)2
the difference 0-2
the sum 0+(-2)
and so on
All these things end up being the same.
It would be more confusing if different symbols were used for each case.

Can -x be regarded as (-1)x?
Yes, in fact this can be used to define multiplication by integers in any group, but we gain nothing as in that context they mean the same thing and -x is shorter.

Subtraction should not exist? I don't know what that means. Clearly it is redundant to have addition, subtraction, additive identity, and additive inverse, but it is convenient and harmless. Commonly (as in many universal algebra books) subtraction is taken as the redundant item. In fact we can take all but subtraction as redundant. For example
0:=c-c
-b:=(c-c)-b=0-b
a+b:=a-((c-c)-b)=a-(-b)
(:= means is defined by)

As far as calculators and computers not considering numbers as signed, that is not true. If we have -2 and we perform square we get 4. It is true that many calculators and computers only allow negative numbers to be entered indirectly. A few by subtraction -2=0-2 which is inconvenient, those with additive inverse buttons lose little by not having a negative number button. The other issue is operations have implied parentheses. If the implied parentheses are not where you would like add explicit parentheses. I would not be surprised if some people are surprised that in latex 12^34 produces
$$12^34$$
and we need to enter 12^{34} to produce $$12^{34}$$.

10. Aug 5, 2013

### coolul007

-x is different than -2, a signed number. a variable is always treated as (-1)(x), my objection is that -2 is not treated as a signed number, but a negated number. there is a difference.
If you have one of the newer TI84 calculators -2^2 = -4, try it you'll see.

11. Aug 5, 2013

### lurflurf

I would be interested if you know any books with something like
$$-2^2=4$$
That is a terrible convention. It is highly confusing unless two different symbols are used.

You just need parentheses (-2)^2. The alternatives are worse. Like -2^2=4 or adding a new symbol
@2^2=2
-2^2=-4

12. Aug 5, 2013

### pwsnafu

I remember $-2^2 = -4$ and ${}^- 2^2 = 4$ in textbooks.

13. Aug 5, 2013

### coolul007

The high school algebra books in California use -2^2 = -4 convention. I think mainly because the TI 84 that is used for examples in all of these books get that result. I think it is terrible, too. I have always been taught that numbers carry a sign as a fundamental property of a number.

14. Aug 6, 2013

### lurflurf

Yes some books use two different symbols as I mentioned. If they look similar though it is confusing as in $$-{}^-2^2 = -2^2$$
It does not really matter if we consider -2 one number or a combination as the end result is the same. I see what you are saying though that the notation biases one away from thinking of -2 as a number not as doing something to 2, using a different symbol is not much help with this. One could express the number in base -2 like this
$$-37=10111_{-2}$$
thought treating adjacent numerals as part of a whole is also a convention.

Your calculator problem is not about how the calculator treats the number, but how it treats your input. It knows (-2)^2=4 when you write -2^2 it sees -(2^2)=-4. This is like my above example if a hypothetical calculator did 12^3=8 it would not mean the calculator thinks (12)^3=8 or that 12 is not a single number (though 12 is short for 10+2) it just means it does not know when your number starts. -2 is the same.

15. Aug 6, 2013

### pwsnafu

My Japanese year 7 textbook (1992 p38) gives
$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2) \times (-2) = 16$
$-2^4 = -(2 \times 2 \times 2 \times 2) = -16$
as examples.

I don't have my Australian textbook with me, but I'm sure it used the same convention.

16. Aug 6, 2013

### D H

Staff Emeritus
You are tilting against windmills, coolul007. -2^2 is almost universally interpreted as -(2^2), not (-2)^2. About the only thing that gets it wrong and calculate -2^2 as being 4 is Microsoft Excel. However, Excel would also calculate -x^2 as being equal to x^2.

With regard to your earlier comment,
The additive inverse and subtraction are very closely related. That 0-2=-2 is the definition of -2. The negative integers are the completion of the natural numbers under subtraction.

Look at the very phrase "negative number". It quite literally means "something that isn't a number". (Similarly, the phrases "irrational number" and "imaginary number" have a rather pejorative connotation. "Irrational" = doesn't make a bit of sense, "imaginary" = whimsical or fictional.)

We humans have been counting for tens of thousands of years (e.g., see http://en.wikipedia.org/wiki/Ishango_bone). The development of zero as a number and consequently the negative integers as numbers was rather recent compared to that tens of thousands of years history of counting. The Babylonians (and others) used zero as a placeholder (e.g., 9007) but did not have a concept of zero as a number in and of itself. The *only* civilization that properly came up with the concept of zero as a number, and hence things like -1 and -9007 as numbers, was the (Asian) Indian civilization around 500 AD. While those Indian arithmeticians tried hard to hold their concepts as a religious secret, their mathematical inventions eventually did escape those religious confines to spread to Arabia and then to Europe.

The things that you think of as numbers, the reals, are a very recent invention compared to the long history of counting.

17. Aug 6, 2013

### Atran

IF -(ab)=(-a)b=a(-b) AND -a=(-1)a THEN I conclude that the minus sign (as an operator) can be moved between the factors of one term. Consequently say -5*-3 = -(-5)*3= 5*3 = 15. What is then the proof that an even number of a negative number always produces a positive number?

Btw, is there a book/source which explains this operator (with proofs)?

18. Aug 6, 2013

### coolul007

My comments are strictly based on keeping the purity of a numbering system. It must follow certain rules and maintain those rules throughout. The concept of additive inverse requires only addition of a number signed opposite of each other. This then translates to a number line. An additive inverse is to numbers equidistant, in opposite directions, from zero on a number line. In real life we use subtraction as it was taught to us first as a convenience. However, in Algebra we should use the fact that numbers carry a sign, and use the additive inverse for our computations. Just where I'm coming from...

19. Aug 6, 2013

### D H

Staff Emeritus
Let's start with the definition of of the additive inverse of some quantity a. -a is *not* defined using multiplication by -1. The additive inverse of a, -a, is the unique quantity, if it exists, such that a+(-a)=(-a)+a=0. 0 is the unique quantity, if it exists, such that a+0=0+a=a for all elements a of the set in question.

Multiplication by -1 is not involved in the definition of the additive inverse. You can instead prove that -a = (-1)*a *if* multiplication is defined and if multiplication has an identity element. Addition and multiplication, along with the additive identity 0 and multiplicative identity 1 are of course well defined for the real numbers. I'll leave the proof that -a=(1)*a as an exercise for you. Try doing so. We'll be glad to help if you run into a brick wall.

I presume you mean the product of an even number of negative numbers is positive. You can prove this recursively. I will once again leave this as an exercise for you (but don't be afraid to ask for help.)

Here's a hint: Start by proving that (-a1)*(-a2)=a1*a2 as your base case. Then prove that if (-a1)*(-a2)*···*(-a2n)=a1*a2*···*a2n then (-a1)*(-a2)*···*(-a2n)*(-a2n+1)*(-a2n+2)=a1*a2*···*a2n*a2n+1*a2n+2.

20. Aug 6, 2013

### lurflurf

This has nothing to do to with the number system. Considering -2^2=4 does not mean we think (-2)^2=4 or that - is not part of -2. It just means that we think -2^2 means -(2^2) and that (-2)^4 should be written (-2)^4. As I wrote above algebra books like to define 3 operations
a 0-ary or nonary operation 0
an 1-ary or unary operation -
an 2-ary or binary operation +
we could also define only the binary operation -
It is no problem to use both together as they are equivalent. Sure - is used (either exactly the same one or variations so similar as to encourage indifference) to indicate that a number is negative, the binary operation of subtraction, or the unary operation of additive inverse. It is reasonable to use the same symbol as the three concepts are so similar.

21. Aug 8, 2013

### Atran

(-1)*a+a=a(-1 + 1)=0 and since the only number added to a to produce zero is its additive inverse, then -a=(-1)*a.

22. Aug 11, 2013

### Atran

What is the order of the operator (minus sign) then? Obviously exponentiation precedes it, and so does multiplication and division?

23. Aug 11, 2013

### coolul007

That's only if you treat the minus sign as an operation and not as the sign of a number.

24. Aug 12, 2013

### Atran

Well: "We have −3 × 4 − 5 + (−3) = −(3 × 4) − 5 + (−3) = −12 − 5 − 3 = −20. Note that
we have recognized that 3 × 4 takes precedence over the − signs." Source: http://rutherglen.science.mq.edu.au/wchen/lnemfolder/em01-ba.pdf [Broken]

Last edited by a moderator: May 6, 2017