Why ##a^0=1##?

[Roberto Pavani]

Equation 1 has no solution: $2^x > 0$ for all $x \in \mathbb{R}$. The only "value" satisfying $2^x = 0$ is $x = -\infty$, which is not a real number, and certainly cannot equal $0$.

 

[weirdoguy]

I decided to solve following equations:

Where in the world did you even get this from?

 

[sophiecentaur]

I decided to solve following equations:

1) Let 2^x =0

Your reality check should kick in here, at this stage. Your equation 1) has to be dodgy. We've all agreed that 2⁰ = 1 and all the various 'justifications / proofs are up there.
Any magnitude less than 1 has to be 2 raised to the power of a number less than zero (a negative number). Smaller would be more negative. Think of a log scale graph (just to get pictorial); smaller numbers need more and more negative powers of 2, which means further and further left along the X axis. Your piece of paper would need to be infinitely wide, to the left.
If the idea came from someone else than check before you believe anything else they say. If it came from you then just give your head a wobble. and 'Look before you leap.'.etc.

 

[Mike_bb]

If you have a number and you choose not to multiply it by anything, should it turn into a zero? Or should it stay the same?

This seems logical to me.

But after rereading, your second post confused me:

Exactly. So you want "multiply by no a's" to be "multiply by 1".

How is it possible "to multiply number by no a's"? I understand when $1$ is at the beginning ($1*a*a..a$). But as I understood you meant $a*a*a*a*...*1$. Give me more details please. Thanks.

 

[weirdoguy]

I understand when 1 is at the beginning (1∗a∗a..a). But as I understood you meant a∗a∗a∗a∗...∗1.

What's the difference? Multiplication is commutative, you can write 1 wherever you want.

My thought: you are overcomplicating things.

 

[Ibix]

How is it possible "to multiply number by no a's"?

You just don't do anything. Which you want to be the same as multiplying by 1.

 

[Mike_bb]

I can't understand how does it work if $x*x*x*x^0$?

You just don't do anything. Which you want to be the same as multiplying by 1.

• x³ = 1 × x × x × x
• x² = 1 × x × x
• x¹ = 1 × x
• x⁰ = 1 (we multiply by x zero times, leaving just the starting 1)

 

[Ibix]

I don't understand your problem. You've stated correctly in your bullets that $x^0=1$, so you know that $x\cdot x \cdot x\cdot x^0=x^3$. What's the problem?

 

[Mike_bb]

I don't understand your problem. You've stated correctly in your bullets that $x^0=1$, so you know that $x\cdot x \cdot x\cdot x^0=x^3$. What's the problem?

When you wrote post#2 you probably meant that "to multiply number by no a's" = $a*a*a*a^0$. I can't understand how did you infer that $a^0=1$ from this expression?

 

[Ibix]

When you wrote post#2 you probably meant that "to multiply number by no a's" = $a*a*a*a^0$. I can't understand how did you infer that $a^0=1$ from this expression?

The same way you did in your bullets. Try $a\cdot a\cdot a\cdot a^n$ with $n>0$ and reduce $n$ to zero.

 

[Mike_bb]

The same way you did in your bullets. Try $a\cdot a\cdot a\cdot a^n$ with $n>0$ and reduce $n$ to zero.

I don't understand again.

$a\cdot a\cdot a\cdot a^2$
$a\cdot a\cdot a\cdot a^1$
$a\cdot a\cdot a\cdot a^0$

What is the next step?

 

[Ibix]

Well, the first one is $a\cdot a\cdot a\cdot a\cdot a$. So what's the third one?

Or alternatively, note that the third one is $a^{3+0}$, which had better equal $a^3$, unless you want $n+0\neq n$.

We're re-treading stuff we've already discussed here.

 

[sophiecentaur]

This is getting to be a very boring thread. It's as if someone thinks it could be possible to prove this stuff actually wrong!

 

[Roberto Pavani]

x== 42

 

[Mike_bb]

Well, the first one is $a\cdot a\cdot a\cdot a\cdot a$. So what's the third one?

Or alternatively, note that the third one is $a^{3+0}$, which had better equal $a^3$, unless you want $n+0\neq n$.

We're re-treading stuff we've already discussed here.

Ah, that's what you're talking about.

 

[Mike_bb]

For $a^0=1$ we have symmetry around $a^0$:
$\frac{1}{a^{-1}} = a$
$\frac{a^1}{a^{-1}} = a \cdot a$


For $a^0\neq1$ we don't have symmetry then it's wrong case:
$\frac{a^0}{a^{-1}} = a^0 \cdot a$
$\frac{a^1}{a^{-1}} = a \cdot a$

Thus, $a^0=1$.

 

[jedishrfu]

Trying to understand why $a^0 = 1$ by using the positive-integer interpretation $a \times a \times a \times \cdots$ is the wrong approach.

Everything in mathematics is either a definition or a theorem.

For positive integers (n), exponentiation is defined by

$a^n = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ factors}}.$

This definition does not tell us what (a^0) should be, since there are no factors to multiply.

To extend exponentiation to all integers, we require the exponent laws to continue holding:

$a^m a^n = a^{m+n}$ and $\frac{a^m}{a^n}=a^{m-n}.$

Using these rules, $\frac{a^1}{a}=a^{1-1}=a^0.$

But the left side is simply $\frac{a}{a}=1.$

Therefore $a^0=1.$

Likewise,

$a^{-n}=\frac{1}{a^n}.$

One intuitive way to see this is to look at the progression

$a^3,\ a^2,\ a^1,\ a^0,\ a^{-1},\ a^{-2},\ldots$

where each step is obtained by dividing the prior term by (a).

For (a=2), this gives: $8,\ 4,\ 2,\ 1,\ \frac12,\ \frac14,\ \frac18,\ldots$

So (a^0=1) is exactly the value needed to keep the pattern and exponent laws consistent.

While intuition may guide us, intuition is not the way of math. It is by definition and theorems that we develop a system of math that preserves the algebraic rules of exponentiation.

The same notion inspired the definitions when n is a real number and later when n is a complex number.

 

[Mike_bb]

Trying to understand why $a^0 = 1$ by using the positive-integer interpretation $a \times a \times a \times \cdots$ is the wrong approach.

We can use empty product to define $a^0=1$.

 

[Mark44]

For $a^0=1$ we have symmetry around $a^0$:
$\frac{1}{a^{-1}} = a$
$\frac{a^1}{a^{-1}} = a \cdot a$


For $a^0\neq1$ we don't have symmetry then it's wrong case:
$\frac{a^0}{a^{-1}} = a^0 \cdot a$
$\frac{a^1}{a^{-1}} = a \cdot a$

Thus, $a^0=1$.

I don't follow what you're trying to do here, in particular by what you mean by "wrong case."

 

[MidgetDwarf]

When you wrote post#2 you probably meant that "to multiply number by no a's" = $a*a*a*a^0$. I can't understand how did you infer that $a^0=1$ from this expression?

a^0=1 was defined this way. Nothing more to it.

 

[sophiecentaur]

a^0=1 was defined this way. Nothing more to it.

Hardly an arbitrary choice; for the function to be continuous, the value at zero has to be 1.

 

[sophiecentaur]

Hardly an arbitrary choice; for the function to be continuous, the value at zero has to be 1.

Or, more accurately, The limiting value for y as x approaches zero is one.

 

[Mike_bb]

Hardly an arbitrary choice; for the function to be continuous, the value at zero has to be 1.

This approach doesn't explain why $a^0=1$.

 

[sophiecentaur]

This approach doesn't explain why $a^0=1$.

You want an arm waving reason? Maths isn't an arm waving topic. In the late 60s, I did a Maths Analysis course at Uni. There were ten (or maybe twenty) hourly lectures and it was scrupulous at every step, from what do we mean by Zero, One , forms of Infinity etc., to hard stuff where I fell asleep. Every step can be tracked back to the first few basic lectures, which was a good justification for everywhere we went, subsequently. I'd have to ask you why you made that comment (or at least the modified comment below.

Or, more accurately, The limiting value for y as x approaches zero is one.

Going into Differential and integral Calculus, we learned about 'limits'; a very powerful concept which allow you to discuss places that you can't actually get to.

 

[Mike_bb]

You want an arm waving reason? Maths isn't an arm waving topic. In the late 60s, I did a Maths Analysis course at Uni. There were ten (or maybe twenty) hourly lectures and it was scrupulous at every step, from what do we mean by Zero, One , forms of Infinity etc., to hard stuff where I fell asleep. Every step can be tracked back to the first few basic lectures, which was a good justification for everywhere we went, subsequently. I'd have to ask you why you made that comment (or at least the modified comment below.

Going into Differential and integral Calculus, we learned about 'limits'; a very powerful concept which allow you to discuss places that you can't actually get to.

I don't blindly trust authorities. Limits are used when we want to check whether function continuous at the point or not but limits don't allow to define values at the point. As was mentioned above $a^0=1$ because $a^{x-x}=1$.

 

[weirdoguy]

but limits don't allow to define values at the point.

They do, as $a^x$ clearly shows.

I don't blindly trust authorities.

But you blindly run around the most simple problem with the most simple solution.

 

[Mike_bb]

They do, as $a^x$ clearly shows.

We don't know nothing about $a^x$ before we define $a^x$ for $x=0$ and $x<0$.

 

[Roberto Pavani]

We don't know nothing about ax before we define ax for x=0 and x<0.

Please define a and x

 

[Mike_bb]

Please define a and x

I like this definition (on Wiki article).

 

[sophiecentaur]

We don't know nothing about $a^x$ before we define $a^x$ for $x=0$ and $x<0$.

If you want to have a valid opinion or objection to this stuff then you need to do what I did (or read and follow through a good book. If not then you have to accept some "authority". The concept of a limit is more than just your single example. Neither of us knows enough to do more than accept most of the Maths we use. For example, do you know the mechanics of deriving dy/dx for y=cos(x) and where the 'limit' is involved (plus some elementary trig identities). Look it up; Google AI will help you there.