Is 0^0 indeterminate?

[Realintruder]

When solving for 0^0 the question remains what number when multiplyed by 0^1 equals 0. The answer is any number.

Just as 1^0 is the number when multyplied by the base 1 equals 1^1, the answer is only 1.<br><br>

 

Or more conscisely what does 0^(1-1)=0^0 equal. The answer is (0^1)/(0^1)=(0/0)=0^0 which is not 1.

 

The teaching in schools all across the world that 0^0=1 is a conspiracy.

Edited February 1, 2015 by Realintruder

[John]

This is not a conspiracy. Rather, it's simply a useful convention. 0⁰ can take multiple values (or be left undefined/indeterminate) depending on the context. The most generally useful value, however, is 1. Here are a couple of explanations why:

http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

[MomentTheory]

I feel the idea of zero, is like the god of mathematics, you either believe in it or you don't. Personally, I think zero is simply a reference point for the unknown or balance because by default of equality, mathematics demands symmetry.

 

Even when thinking about any topic, to try and say I have NOTHING of SOMETHING, you still must think of the something, and that in itself has value.

 

With that logic you can see how 0^0 = 1, the value originating from in the idea of the operator "^", the operator outside of the 0, inserts value. Since the value is unknown, we say 1. I can see how one might feel this is a "conspiracy".

[Realintruder]

If we allow b to equal 1 and 0^b=0^1=0=0^1

Then

0^(b-1)=0^0=0^(1-1)=(0^1)/(0^1)=0/0

Let's hear your argument against that.

Likewise if 0^0=1 then (0^0)^-1 should equal 1 since 1^(-1) is equal to 1. But it does not

 

(0^0)^-1=(0^-1)/(0^-1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0

Edited February 2, 2015 by Realintruder

[swansont]

!

Moderator Note

Let's leave discussion of conspiracy out of this. Continue with the math (as in, don't respond to this modnote)

[fiveworlds]

 

anything^0=0

 

[Strange]

anything^0=0

 

Er, no: http://www.wolframalpha.com/input/?i=\pi^0

[fiveworlds]

mmm.

2^3=(2)(2)(2) add three 2s

2^2=(2)(2) add 2 2s

2^1=(2) add 1 2

2^0=0 because you add no 2s to multiply by

[Strange]

mmm.

2^3=(2)(2)(2) add three 2s

 

That is multiply, not add. (I worry about you, some days.)

[fiveworlds]

That is multiply, not add. (I worry about you, some days.)

 

This is the function.

<script>
var number=2;
var exponent=0;
function isFloat(n) {return n === +n && n !== (n|0);}
function isInteger(n) {return n === +n && n === (n|0);}
var logic=false;
var fn=false;
var fc=false;
var fx=number;
if(fx<0){fx=(fx*(-1));fn=true;}
var fy=exponent;
if(fy<0){fy=(fy*(-1));fc=true;}
if((isFloat(fx)==true||isInteger(fx)==true)&&(isFloat(fx)==true||isInteger(fx)==true)){logic=true;}
var fz=0;
var ft=false;
var count=0;
if((isFloat(fz)==true||isInteger(fz)==true)&&(isFloat(count)==true||isInteger(count)==true)&&(logic!=false)){logic=true;}
while((count < fy) && (logic=true))
{
if(fz>1){fz=fz*fx;}
if((fz<1) && (fz>-1) && (ft != true)){fz=fx;ft=true;}
count=count+1;
}
if(fn==true){fz=(fz*(-1));}
if(logic==false){document.write("syntax error")}
if((logic==true)&&(fc==true)){fz=(1/fz);document.write(fz);}
else{document.write(fz);}

</script>

Edited February 2, 2015 by fiveworlds

[Realintruder]

I meant for this to go on a seperate block but the software would not let me. I cannot edit it as it has timed out that I can no longer edit recent posts on this page, so I must repost it!:<br><br>

 

Likewise if 0^0=1 then (0^0)^-1 should equal 1 since 1^(-1) is equal to 1. But it does not

 

 

(0^0)^-1=(0^-1)/(0^-1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0

[imatfaal]

mmm.

2^3=(2)(2)(2) add three 2s

2^2=(2)(2) add 2 2s

2^1=(2) add 1 2

2^0=0 because you add no 2s to multiply by

The point is that if you take your final statement

2⁰=0

and multiply both sides by 2 you would end up with the following

2(2⁰)=2(0)

now we know that 2(2ⁿ) = 2ⁿ⁺¹ so we end up with

2(2⁰) = 2⁰⁺¹ = 2 = 0 ie a contradiction

 

BUT if we define 2^0 =1

then we get

2(2⁰) = 2(1) = 2

[John]

If we allow b to equal 1 and 0^b=0^1=0=0^1

 

Then

 

0^(b-1)=0^0=0^(1-1)=(0^1)/(0^1)=0/0

 

Let's hear your argument against that.

I have no argument against that, as it's a perfectly valid line of reasoning. However, there are other possible interpretations, as discussed in the pages I linked earlier. As I said before, 0⁰ can be considered indeterminate, but there are other possible values, and 0⁰ = 1 is the most generally useful.

 

Likewise if 0^0=1 then (0^0)^-1 should equal 1 since 1^(-1) is equal to 1. But it does not

 

(0^0)^-1=(0^-1)/(0^-1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0

I'm not sure what you're getting at here, but your first step seems wrong, and since 1/0 is undefined regardless, the chain of equalities is invalid. Anyway, using the convention that 0⁰ = 1, if we have (0⁰)^(-1), then we have (0⁰)^(-1) = 0^(0)(-1) = 0⁰ = 1, as expected.

[Greg H.]

Normally, whenever I hit the word "conspiracy" in a post about science or math, I assume the rest of it is quackery, and move on with my day.

 

However, this one interested me enough to actually go looking for information and, John succinctly puts it, it depends on the context. In everday math, 0^0 = 1 is perfectly fine, and keeps things (like the binomial theorem) working as expected. However, in certain scenarios (the source I read lists limiting forms as one such scenario), you can consider it to be undefined.

 

In addition to John's resources, you can also review the findings at https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

[Realintruder]

Not so, as discussed earlier.

 

For if 0^0 indeniably equals 0^(1-1)=(0^1)/(0^1)=0/0

 

Except in practical applications, which only exist in "man made' settings, do we have 0^0=1, otherwise it exists only, as the number when multiplyed by 0 equals 0, such that is exponently less than 0 that is 0.

 

If x is 0 then when:

 

x^0 times x^1 must equal base 0 as well. When x^0 equals 0 that holds true for any number complex or real for x^0.