Questioning 10^0: Does it Equal 1 or 0?

[Deicider]

Homework Statement

the title says it all,its very simple question but i can't get it.

Homework Equations

The Attempt at a Solution

10^2=100 =10*10
10^1=10 = 10*1
10^0=1(?) =10*0=0 ...1?
I never asked anyone this,and never questioned it just followed what teacher said but now i want to some answers : D

 

[Dick]

If you want the laws of exponents to work correctly you need x^0*x^n=x^(0+n)=x^n. Since x^0 multiplied by x^n is x^n, x^0 must be 1.

 

[Deicider]

If you want the laws of exponents to work correctly you need x^0*x^n=x^(0+n)=x^n. Since x^0 multiplied by x^n is x^n, x^0 must be 1.

2^0*2^2 =2^(0+2)=2^2=4
if
2^0=2*0=0 and 2^2=2*2=4
then
0+4=4
isnt the same? i think it works with being zero too.
i know am wrong cause ,obviously, all mathematicians(and non) would agree with you,but with my 'denial' i just want to understand it better and so far i presume that this has to do with "bending" a property of numbers to fit in exponent laws as u said.
or smth like that

 

[Mark44]

2^0*2^2 =2^(0+2)=2^2=4
if
2^0=2*0=0 and 2^2=2*2=4

But 2^0 $\neq$ 0

then
0+4=4
isnt the same? i think it works with being zero too.
i know am wrong cause ,obviously, all mathematicians(and non) would agree with you,but with my 'denial' i just want to understand it better and so far i presume that this has to do with "bending" a property of numbers to fit in exponent laws as u said.
or smth like that

 

[Dick]

2^0*2^2 =2^(0+2)=2^2=4
if
2^0=2*0=0 and 2^2=2*2=4
then
0+4=4
isnt the same? i think it works with being zero too.
i know am wrong cause ,obviously, all mathematicians(and non) would agree with you,but with my 'denial' i just want to understand it better and so far i presume that this has to do with "bending" a property of numbers to fit in exponent laws as u said.
or smth like that

Defining 2^0=0 doesn't work! 2^0*2^2 should be 2^(0+2)=2^2=4. 0*4 is NOT 4. You can't just change '*' into '+'.

 

[Deicider]

i just found a pretty good example:
"""""""""
3^1, 3^2, 3^3, 3^4, ...
3 , 9 , 27 , 81 , ...

So what is the pattern in the bottom sequence? Well, every time you move to the right in the list you multiply by 3, and every time you move to the left in the list you divide by 3. So we could take the bottom sequence and keep going to the left and dividing by 3, and we'd have the sequence that looks like this:

..., 3^-3, 3^-2, 3^-1, 3^0, 3^1, 3^2, 3^3, 3^4, ...

..., 1/27, 1/9, 1/3 , 1 , 3 , 9 , 27 , 81 , ...
"""""""""
as goes beneath ^1 still sucks intuitively : D

 

[elect_eng]

Dick's explanation is correct of course.

If it's still not sinking in, then think of something like 2^x as a function. Plot the function and look at the value of the function at x=0. I've attached a plot to make it easy to see.

 

[sylas]

Try this.

write down 2¹, 2², 2³, 2⁴ and so on. You should get

2, 4, 8, 16, 32, 64, ...

Going backwards, you divide by 2, don't you.

64, 32, 16, 8, 4, 2, ...

So extend it.

64, 32, 16, 8, 4, 2, 1, 0.5, 0.25, 0.125, ...

This helps give the intuition that 2⁰ = 1; it also shows what you expect for negative powers!

Cheers -- sylas

 

[icystrike]

Homework Statement

the title says it all,its very simple question but i can't get it.

Homework Equations

The Attempt at a Solution

10^2=100 =10*10
10^1=10 = 10*1
10^0=1(?) =10*0=0 ...1?
I never asked anyone this,and never questioned it just followed what teacher said but now i want to some answers : D

$$1 = \frac{x}{x} = (x)(x^{-1}) = x^{1+(-1)} = x^{1-1} = x^{0}$$

Conversely you can say that $$x^{0} = 1$$

 

[Post]

Use the division rules of exponents. (x^a)/(x^b) = x^(a-b)

if a = b then a - b = 0, and any number divided by itself is 1. Done

 

[HallsofIvy]

2^0*2^2 =2^(0+2)=2^2=4
if
2^0=2*0=0 and 2^2=2*2=4
then
0+4=4

But you switched from * to +! 0*4= 0, not 4.

isnt the same? i think it works with being zero too.
i know am wrong cause ,obviously, all mathematicians(and non) would agree with you,but with my 'denial' i just want to understand it better and so far i presume that this has to do with "bending" a property of numbers to fit in exponent laws as u said.
or smth like that

No, no bending involved. It only requires recognizing that addition is not multiplication!

 

[Martin Rattigan]

$x^0=1$ may not be something you prove at all.

Before you could prove it you have to know what $x^y$ means. That comes from a definition.

In this case there are a series of definitions for $x^y$, usually starting with the case when $y$ is a natural number (the ones you count with starting 0,1,2) and adding further definitions to cope with negative integers, fractions, complex numbers etc., at each stage making sure that the definition added doesn't conflict with the ones you have already made.

The usual approach to defining $x^y$ when $y$ is a natural number is to define $x^0$ to be $1$ for any number $x$ and then define $x^n$ to be $x^{n-1}x$ for any $n>0$. Having defined the expression for $y=0$, the second part successively defines it for 1, 2, 3 and eventually for all natural numbers.

With this approach, $x^0=1$ exactly because you say so and for no other reason. The proof, such as it, is simply that $x^0=1$ by definition.

Of course there is no harm in speculating on why people define it that way and what the consequences are.

 

[Susanne217]

Homework Statement

the title says it all,its very simple question but i can't get it.

Homework Equations

The Attempt at a Solution

10^2=100 =10*10
10^1=10 = 10*1
10^0=1(?) =10*0=0 ...1?
I never asked anyone this,and never questioned it just followed what teacher said but now i want to some answers : D

I am not trying to be a smart ***, but as I said in another post. This is a pre-Calculus forum which means its suppose to be problems which doesn't involve the usage of Real Analysis.

You just to accept that that is the definition. If not grab a Real Analysis book and its explained to you there :)