This example will prove that if x

^{2}+ y

^{2}= z

^{2}then (v

^{2}- 1)z

^{2}- 2xy = 0(mod x+y+vz), where v can be any value.

Step one: Start with a truth

1. x+y+vz = x+y+vz

A lof of my research begins with an identity, which of course is true. For me a truth is something that does not change. Which means it must be true no matter what.

Proceed using logical steps, where some might prefer to say, valid mathematical steps.

2. x+y+vz = 0(mod x+y+vz)

(If "mod" is unfamiliar to you or you have questions about my usage of it, please check out this post.)

That simply states that x+y+vz has itself as a factor, and is actually a truth from which one could start, but I like to begin with the explicit identity, when I'm explaining in a lot of detail.

3. x+y = -vz (mod x+y+v)

Subtracted vz mod x+y+vz from both sides.

4. x

^{2}+ 2xy + y

^{2}= v

^{2}z

^{2}(mod x+y+vz)

Here I've just continued with basic algebra, squaring both sides. If you're worried you can do it all explicitly. Now it's time to introduce what I call a conditional.

5. Let x

^{2}+ y

^{2}= z

^{2}and subtract from 4.

It is, unlike the identity with which we started, not always true. So now things remain true under certain conditions.

6. 2xy = (v

^{2}- 1)z

^{2}(mod x+y+vz)

Simplifying a bit, more for presentation.

7. (v

^{2}- 1)z

^{2}- 2xy = 0(mod x+y+vz)

Proof complete.

This type of result is true with the condition that x

^{2}+ y

^{2}= z

^{2}, so it's like, if that is true then

(v

^{2}- 1)z

^{2}- 2xy = 0(mod x+y+vz) must be true.

That conditional result is what I call a conditional residue.

This simple example shows how you begin with a truth, and connect truth to truth, as well as use a conditional to find what must be true when that condition is met.

Notice v is a perfectly independent variable.

As long as x, y and z fit the conditional, for instance, x = 3, y = 4, z = 5, then v can take any value.

For example: (v

^{2}- 1)25 - 2(3)(4) = 0(mod 3+4+5v)

Which is just: 25v

^{2}- 49 = 0(mod 5v + 7), which can be seen to be true by inspection.

For a second example, let v=1, then you have -2xy = 0(mod x+y+z) is true if x

^{2}+ y

^{2}= z

^{2}.

With our values: -2(3)(4) = 0( mod 3+4+5), which is -24 = 0 mod 12

So now you know that if x

^{2}+ y

^{2}= z

^{2 }then 2xy has x+y+z as a factor.

As v is completely free there are an infinity of such relations available. Notice you can also probe behavior of x, y and z by what v you pick, so it can be an analysis tool.

The conditional residue contains all information about the conditional itself, which allows you to probe behavior of the conditional through it.

The purpose of this post is to step through showing a truth, logical steps, which are also valid mathematical steps, to get to a conclusion which must be true.

Once you understand the basics, you can handle far more complex examples and understand why you have absolute proof without any possibility of doubt.

Each of the steps above, is both logical, and a valid mathematical step. I call those steps, linkages.

For example, the linkage from x+y+vz = 0(mod x+y+vz) to x+y = -vz(mod x+y+vz), is the logical step of subtracting vz mod x+y+vz from both sides. Notice that is a valid mathematical step.

One of those things I rarely do with my own arguments is state when a proof is complete, but do so at times when explaining in detail.

You may notice that the argument was perfect at each step.

Correct mathematical arguments have the property of being correct at each step.

James Harris

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