### Saturday, April 08, 2006

## Square roots, two values

One of the more bizarre aspects of continuing resistance to my research are arguments that I keep seeing where people rely on the

But, um, people, if you have sqrt(4), sure, convention can be that the answer is 2, but reality is that -2 works as well.

There are TWO solutions. The square root of 4 is not just 2, but is 2 and -2 as EITHER WORKS.

Convention may be one thing, but the mathematics says that the function has two answers.

Some people a while back didn't like two answers for functions so they proclaimed that functions should have only one value. So?

It's a convention, a human thing, just something people decided to do, to claim that the square root function has only one value. The other value is usually just ignored.

But with my research both values have impact, so you can get wrong results by just throwing away the other value, and sometimes it takes me a while to notice when some poster does that, but whether or not I catch you it doesn't change what's mathematically correct.

value still exists!!!

I find it to be one of the more surprising aspects of all of this that so many of you act as if it is beyond your comprehension that despire the decision of some people over a hundred years ago that functions should only have one value that the square root of a number will STILL be two answers.

Can this issue die yet? Or am I going to keep seeing people claiming to refute my work by relying on taking only a single value for the square root?

[A reply to someone who asked for an example of someone formally trained in mathematics who would deny that there are two numbers that can be the square root of a number.]

Rick Decker and W. Dale Hall are in a thread on sci.math where they are pushing what they claim is a counterexample to my simple analysis of

7*C(x) = (f(x) + 7)(g(x) + 1)

where C(x) is a polynomial and where a given requirement is that f(0) = g(0) = 0.

Their functions f(x) and g(x) are two-valued at x=0 by clever use of a square root.

So you have f(0) = 0 or -6.

That's the kind of behavior that I've learned to expect from people fighting my results.

You don't give a damn about what is actually correct.

For those who wonder what the fuss is about, if you do cheat, and ignore the rule that f(0) = g(0) = 0, you can appear to break a proof.

The proof is that since the distributive property doesn't care about the value of what it's multiplying, since it's just the statement that a multiplication distributes within a group

e.g. a*(f(x) + b) = a*f(x) + a*b

is true regardless of the value of f(x), you can, when there is some question about how a factor, like 7 in

7*C(x) = (f(x) + 7)(g(x) + 1)

distributes, you can take a convenience value—because the distributive property doesn't care about the value of what's being multiplied—and check there.

When f(x) and g(x) are both 0 you have a convenience value, but they have to just be 0, and not multi-valued at that point.

That shows that you have this general result when factoring a polynomial in this odd way that 7 will distribute to only one factor giving you

(f(x) + 7)

which is why the 7 is actually visible there.

But you can find examples in the ring of algebraic integers where that conclusion shows a problem with the ring, which is what people like Decker and Hall are trying to deny.

But the result is mathematically correct, so they have to cheat.

In this case, they cheat by using the square root as single-valued.

To me the reality that people cheat over and over again to hide from correct mathematical results is a community problem. Your community does not hold its members accountable.

You people get upset about a result, and there's nothing anyone can do to force you to follow what's mathematically correct, as you're like a gaggle of children with no rule by authority, or deference to what is mathematically correct.

You care only about how you FEEL about a solution, not what is proven.

[A reply to someone who said that the use of the absolute value of the square root eliminates every ambiguity.]

Sorry, but if what you are trying could work, I'd go right along with it.

It just does not work.

If somehow, someway you could use the absolute value to force the square root to return a single value in a mathematical sense that mattered then that would be one thing.

But it is just the same as saying—take the positive root only.

Consider this, aliens on planet Contrary say that the absolute value is the negative!!!

Can you say they're wrong?

Does the mathematics on planet Contrary really differ from that on planet Earth?

Or is mathematics an absolute?

**convention**that square roots return a single value, to claim counterexamples.But, um, people, if you have sqrt(4), sure, convention can be that the answer is 2, but reality is that -2 works as well.

There are TWO solutions. The square root of 4 is not just 2, but is 2 and -2 as EITHER WORKS.

Convention may be one thing, but the mathematics says that the function has two answers.

Some people a while back didn't like two answers for functions so they proclaimed that functions should have only one value. So?

It's a convention, a human thing, just something people decided to do, to claim that the square root function has only one value. The other value is usually just ignored.

But with my research both values have impact, so you can get wrong results by just throwing away the other value, and sometimes it takes me a while to notice when some poster does that, but whether or not I catch you it doesn't change what's mathematically correct.

- The square root function by convention is said to return only one value, where people take the positive, by convention, but that doesn't destroy the other solution! So -2 is still an answer to sqrt(4) no matter what people say.
- My research pushes mathematical ideas to their limits in this area, and the error of throwing away the other solution for the square root has an impact.
- Posters who rely on the error which I am pointing out to claim counterexamples are just being deliberately obtuse.

value still exists!!!

I find it to be one of the more surprising aspects of all of this that so many of you act as if it is beyond your comprehension that despire the decision of some people over a hundred years ago that functions should only have one value that the square root of a number will STILL be two answers.

Can this issue die yet? Or am I going to keep seeing people claiming to refute my work by relying on taking only a single value for the square root?

[A reply to someone who asked for an example of someone formally trained in mathematics who would deny that there are two numbers that can be the square root of a number.]

Rick Decker and W. Dale Hall are in a thread on sci.math where they are pushing what they claim is a counterexample to my simple analysis of

7*C(x) = (f(x) + 7)(g(x) + 1)

where C(x) is a polynomial and where a given requirement is that f(0) = g(0) = 0.

Their functions f(x) and g(x) are two-valued at x=0 by clever use of a square root.

So you have f(0) = 0 or -6.

That's the kind of behavior that I've learned to expect from people fighting my results.

You don't give a damn about what is actually correct.

For those who wonder what the fuss is about, if you do cheat, and ignore the rule that f(0) = g(0) = 0, you can appear to break a proof.

The proof is that since the distributive property doesn't care about the value of what it's multiplying, since it's just the statement that a multiplication distributes within a group

e.g. a*(f(x) + b) = a*f(x) + a*b

is true regardless of the value of f(x), you can, when there is some question about how a factor, like 7 in

7*C(x) = (f(x) + 7)(g(x) + 1)

distributes, you can take a convenience value—because the distributive property doesn't care about the value of what's being multiplied—and check there.

When f(x) and g(x) are both 0 you have a convenience value, but they have to just be 0, and not multi-valued at that point.

That shows that you have this general result when factoring a polynomial in this odd way that 7 will distribute to only one factor giving you

(f(x) + 7)

which is why the 7 is actually visible there.

But you can find examples in the ring of algebraic integers where that conclusion shows a problem with the ring, which is what people like Decker and Hall are trying to deny.

But the result is mathematically correct, so they have to cheat.

In this case, they cheat by using the square root as single-valued.

To me the reality that people cheat over and over again to hide from correct mathematical results is a community problem. Your community does not hold its members accountable.

You people get upset about a result, and there's nothing anyone can do to force you to follow what's mathematically correct, as you're like a gaggle of children with no rule by authority, or deference to what is mathematically correct.

You care only about how you FEEL about a solution, not what is proven.

[A reply to someone who said that the use of the absolute value of the square root eliminates every ambiguity.]

Sorry, but if what you are trying could work, I'd go right along with it.

It just does not work.

If somehow, someway you could use the absolute value to force the square root to return a single value in a mathematical sense that mattered then that would be one thing.

But it is just the same as saying—take the positive root only.

Consider this, aliens on planet Contrary say that the absolute value is the negative!!!

Can you say they're wrong?

Does the mathematics on planet Contrary really differ from that on planet Earth?

Or is mathematics an absolute?