```Date: Dec 1, 2012 2:30 PM
Author: Paul A. Tanner III
Subject: In "square root of -1", should we say "minus 1" or "negative 1"?

In "Re: Some important demonstrations on negative numbers"http://mathforum.org/kb/message.jspa?messageID=7930782Jonathan Crabtree wrote:....> Marcus du Sautoy and other professors keep talking about the square root of MINUS one when they mean the square root of NEGATIVE one. People reading this probably have as well.>> People state the plus and minus sign when they mean the positive and negative sign.I think that this is not correct when professional mathematicians and some others say "minus". The clear convention that mathematicians use (and we see this in AMS articles and many other places) is that "minus" is used to be a shorter and easier way of saying "the additive inverse of" regardless of whether the set under discussion is an ordered set, while "negative" is reserved for "the additive inverse of" only when the set under discussion is an ordered set, specifically an ordered ring.I understand that the convention that has evolved is this below, and for the following reasons: The terms "positive" and "negative" have been conventionally defined only for sets in which the entire set is under an order that preserves the algebra, where "negative" means "less than 0" and "positive" means "greater than 0".In a ring, additive inverses are more general than negatives, by the convention that we need a ring to be more restricted axiomatically to be not just a ring but at least a partially ordered ring, where under an ordering "negative" is defined to mean "less than 0" and "positive" is defined to mean "greater than 0". And therefore if we are to have a shorter and easier way of saying "the additive inverse of", the convention that has evolved (saying "minus x" instead of "negative x" as the defined easier and shorter way of saying it) is the right one. That is, it would be wrong to say "negative x" when x is neither greater than nor less than 0. That is, since it applies in both ordered and non-ordered sets, "minus" is the much more general term to mean "the additive inverse of" than the term "negative", which holds as a legitimate shortcut for "the additive inverse of" only in ordered sets. That is, more generally, since some non-ordered sets contain ordered subsets by which it would be OK to say "negative x" for those elements in those ordered subsets, it's still best to say "minus x" for all elements in non-ordered sets to avoid confusion as to when we can and cannot say "negative x" in non-ordered sets.An example of these last points is -i. Element i is neither less than nor greater than 0, and so it would be inconsistent with the conventional definitions for "negative" and "positive" as "less than 0" and "greater than 0" to say "negative i". And so, since the term "minus" was an easy and handy way to have a shortcut term for "the additive inverse of" for all elements regardless of whether the sets are ordered, "minus i" evolved to be that shortcut term for -i. Thus -1 in "the square root of -1" in the complex numbers is best said as "minus 1" instead of "negative 1" to try to avoid confusion. (Yes, since -1 is a real number and is a member of the ordered set called the reals, we can say "negative 1". But still, since being able to denote -x as "negative x" holds only for the real subset of the complexes, it's still best to use one universal term "minus" to cover all the additive inverses in the entire non-ordered set of complexes. At least that is the convention that has evolved.)Final note: As for such theorems as (-a)(-b) = ab for all a,b in a ring (including the special case of (-a)(-a) = aa = a^2), how are they always true for all elements in rings including the non-ordered ones including complex numbers? Simple. These theorems are about additive inverses, regardless of whether the sets are ordered. (This means that this theorem is not in conflict with such as (i)(i) = -1, which we find in the complex numbers, since we are using an element i in the non-ordered complexes to say this, not an element in the ordered subset of the reals to say this. That is, since we use an element outside of the ordered set, the equality (i)(i) = -1 does not contradict the theorem that in an ordered set, the square of any element in that ordered set is nonnegative.)
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