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."