Does 0 = 1?

[Obnoxious]

Because if you start out with the equation:

1^1 = 1^0

(1)Log 1 = (0)Log 1

1 = 0

Or did I do something naughty?

[Nevermore]

You have already posted this question, and gotten a perfectly good answer.

[Obnoxious]

Humor me, I seem to have forgotten.

[Dapthar]

Humor me, I seem to have forgotten.

[math]log[/math][math] 1=0[/math], so dividing by [math]log[/math][math]1[/math] is invalid.

[ed84c]

yes its that calssic divide by 0 and you prove that one number = the same as another.

 

You can use a similar method in algebra to prove 1=2. Again, in reality you are dividing by 0.

[dan19_83]

What about the rule that any number to the power of zero is always equal to one?

1^1=1^0

 

simply becomes:

1=1

[Dave]

I think the idea was to see what was wrong with his "proof" of sorts.

[Asimov Pupil]

[math]log1=0[/math]

 

therefore [math]0=0[/math]

 

or am i missing something?

[Callipygous]

What about the rule that any number to the power of zero is always equal to one?

1^1=1^0

 

simply becomes:

1=1

 

 

1^37billion kajillion=1^3

 

so 37billion kajillion = 3?

[Johnny5]

Because if you start out with the equation:

1^1 = 1^0

(1)Log 1 = (0)Log 1

1 = 0

Or did I do something naughty?

 

Let it be the case that

 

[math] 1^1 = 1^0 [/math]

 

Take the natural logarith of both sides of the statement above' date=' so the following statement has the same truth value as the previous one:

 

[math'] ln(1^1) = ln(1^0) [/math]

 

Here is a brief tutorial on logarithms in general:

 

Hyperphysics on logarithms

 

Notice a little ways down, they say that log to the base b of N to the y, is equal to y times log to the base b of N, look for:

 

[math] log_b N^y = y log_b N [/math]

 

Using natural logs, that is written as follows:

 

[math] ln N^y = y ln N [/math]

 

Where the base b, has been chosen to be e=2.71828... , and is now implicit instead of explicit.

 

continuing on...

 

[math] ln(1^1) = ln(1^0) [/math]

 

Using the natural log rule it follows that:

 

[math] 1 ln 1 = 0* ln(1) [/math]

 

The natural logarithm of 1, is equal to zero, that is:

 

[math] ln 1 = 0 [/math]

 

Therefore:

 

[math] 1 *0 = 0*0 [/math]

 

And it is provable, using the axioms of the real number system, that zero times anything equals zero, therefore:

 

[math] 0 = 0 [/math]

 

And there is no contradiction.

 

The following steps in your original argument was invalid:

 

(1)Log 1 = (0)Log 1

1 = 0

 

And the reason is that Log 1=0, and you cannot divide by zero, because that will contradict the field axioms.

 

Regards

 

If you look down at the link below you will see that log to any base of 1 is equal to zero.

 

Log 1 = 0 regardless of base

 

I usually do most things using the natural log, but you can go from a logarithm of any base, to natural logs using the following relation:

 

[math] log_a x = \frac{ln x}{ln a} [/math]

 

and Log is usually reserved for log to the base ten.

 

Hence:

 

[math] log_{10} x = Log x = \frac{ln x}{ln 10} [/math]

 

Therefore:

 

[math] ln 10 log_{10} x = Log x = ln x [/math]

 

I am presuming you mean log to the base ten, since you wrote Log.

 

Multiply both sides of the statement above by ln 10, to get:

 

[math] ln(10) * (1)Log 1 = ln (10) * (0)Log 1 [/math]

 

Therefore:

 

[math] (1) ln 1 = (0) ln 1 [/math]

 

I only converted from base ten to base e, because I am used to using the natural log, but you don't have to do that. Nonetheless, from memory I know that ln 1 = 0, and you cannot divide by zero, your step here

 

(1)Log 1 = (0)Log 1

1 = 0

 

is against the field axioms; Log 1 = 0 as well.

 

Regards