Science Forums 2

[Don Blazys]

To:Ughaibu

0/0=N does indeed tell us that N can be any non-negative integer.

Therefore, it is called "indeterminate" because if there is no further information on how we got 0/0, then we can't determine it's exact value.

It's only use is in making general statements such as: "One raised to any power equals one," or in mathematical symbols: 1^(0/0)=1^N=1.

However, this thread is not about "indeterminate forms". It's about the proper restrictions that must occur if we are to represent and eliminate common factors correctly.

For instance, if we have the equation:

Ta^x+Tb^y=Tc^z,

where all the variables are positive integers, then how do we eliminate the common factor T so that the equation becomes "co-prime" (Contains no common factor.)?

Well, these days, students are being taught that we should divide each and every T by T, then "cross out" the T's so that they "disappear". Doing so gives us:

(T/T)a^x+(T/T)b^y=(T/T)c^z = a^x+b^y=c^z.

However, this is wrong, because it implies that x, y and z can all be greater than 2 when there is no common factor.

In reality, when there is no common factor, then we must have a "restriction" on either x, y or z so that either:

x={1,2}, y={1,2} or z={1,2}.

Where did we go wrong? Well, we never actually prevented T=1, did we? Preventing T=1 is important, because a true or "non-trivial" common factor is defined as T>1.

Now, watch what happens if we refuse to "cross out" the T's "prematurely" and re-write the co-prime equation:

(T/T)a^x+(T/T)b^y=(T/T)c^z

so that it appears as either:

(T/T)a^x+(T/T)b^y=T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)),

or (in it's factored form):

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2.

Immediately we find that by substituting just one "Blazys term", we eliminated any possibility that T=1. Now, take a good close look at the last three equations. Notice that the first one tells us that T=c is allowable while the next two tell us that before we can allow T=c, we must first let z=1 and z=2, then immediately "cross out" the logarithms themselves. Thus, the last three equations now appear as:

(T/T)a^x+(T/T)b^y=(T/T)c^z,

(T/T)a^x+(T/T)b^y=T(c/T)

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=(T(c/T))^2.

Now, and only now can we allow T=c, or we can simply "cross out" the remaining T's so that the above three equations appear as:

a^x+b^y=c^z,

a^x+b^y=c

and

a^x+b^y=c^2.

Notice that the first of the above three equations is a lie because it implies that if we add together any two co-prime numbers a^x, (x>2) and b^y, (y>2), we might get a third number c^z where z>2.

The other two equations tell us the truth, which is that if we add together any two co-prime numbers a^x, (x>2) and b^y, (y>2), then the exponent of c must be either 1 or 2.

Try it yourself! Add together any two co-prime positive integers under the sun with exponents greater than 2 and you will find that their sum will always have an exponent of either 1 or 2.

Most importantly, notice that "indeterminate forms" such as 0/0 are never ever encountered if we do the algebra correctly and "cross out" or "cancel out" the expressions involving logarithms the very moment that we let z=1 and z=2.

Believe it or not, there are some mathematicians who don't think it's possible to "cross out" or "cancel out" the logarithms at z=1 and z=2.

I think that they are mistaken.

I think that conjuring up "indeterminate forms" that don't even exist is just plain silly.

Not only is this the correct way to represent and eliminate "common factors", but it also shows us that problems such as the "Beal Conjecture" and "Fermat's Last Theorem" would never have existed had mankind learned how to properly represent and eliminate common factors to begin with! Thus, it's quite understandable that many in the math community find this irrefutable result to be "embarrassing".

It is the solution to those supposedly "hard" problems!

Don.