Unresolved "To the power of half"?

Question

I have seen many formula to resolve this but everyone have tried to make it logical,to say square it and multiply two exponents, such as $(x^{\frac{1}{2}})^2 = x^{\frac{1}{2} \cdot 2} = x^1$.

But I am searching an explanation in terms of repeated multiplication: for example, $2^3$ means $2 \cdot 2 \cdot 2$, but how this can be extended to understand why $16^{\frac {1}{2}}$ means $\sqrt{16}$?

Is it impossible to multiply 16 by itself 1/2 times,if possible what is the algebric expression without any variable i mean using only 16 and 1/2?

Answer 1

Based on your question, I assume you are asking not for an algebraic demonstration of why "to the 1/2 power" means "square root", but rather an intuitive explanation of why it should mean that. Here is my best attempt at explaining why that makes sense.

Suppose I ask you the question: What number is halfway between $6$ and $96$?

You might reason as follows: To get from $6$ to $96$, I add $90$. If I want to break that up into two equal steps, I could think of it as adding $45$, then adding $45$ again. The value in the middle would be what I get if I start with $6$ and add $45$ -- in other words, $51$.

That would be a perfectly reasonable explanation and answer if you are only able to think of getting from one number to another by way of addition. But there is another perspective on the question, one that thinks of getting from one number to another by way of multiplication.

That second way of reasoning goes as follows: To get from $6$ to $96$, you multiply by $16$. Now ask yourself:

How do I break the operation "multiply by $16$" into two equal "half-steps"?

A naïve first answer might be to think: "Half of $16$ is $8$, so I multiply by $8$ twice." But it's easy to see that doesn't work: If you start with $6$, multiply by $8$, then multiply by $8$ again, you definitely don't end up at $96$.

What does work is to think:

"If I multiply by $4$, then multiply by $4$ again, the result is the same as just multiplying by $16$ in a single step."

And you can check this: Start with $6$, multiply by $4$ to get $24$, then multiply by $4$ again to get $96$. The number "in the middle" is $24$.

So there are two different types of reasoning involved: Additive reasoning and Multiplicative reasoning:

• According to additive reasoning, "half of $+90$" means $+45$, and the number halfway between $6$ and $96$ is $51$. This is called the arithmetic mean (or just the average) of $6$ and $96$.
• According to multiplicative reasoning, "half of $\times 16$" means $\times 4$, and the number halfway between $6$ and $96$ is $24$. This is called the geometric mean of $6$ and $96$.

To understand why $16^{1/2}$ means "the square root of $16$", you have to reason multiplicatively. As you know, $16^5$ means "multiply $5$ factors of $16$ together". Likewise for any positive whole number $k$, $16^k$ means "multiply $k$ factors of $16$ together." In both cases, we are thinking multiplicatively, not additively. So to make sense of $16^{1/2}$ we should understand it as meaning "Multiplying by $16$ 'half of one time'." As the discussion above explains, the only sensible interpretation of that is that it means "Multiply by $4$".

Answer 2

From the idea of exponentiation as repeated multiplication we notice certain principles such as

$$ b^m\cdot b^n=b^{m+n} $$

in the case of natural numbers. But then we notice that these principles can be extended for fractional exponents if we follow the same rules, such as

$$ 16^{1/2}\cdot 16^{1/2}=16^{1/2\,+\,1/2}=16^1=16$$

so we conclude that this only makes sense if $16^{1/2}=4$.

Following the same rule we can make sense of a zero exponent if we consider examples such as

$$ 5^1\cdot5^{0}=5^{1+0}=5^1=5$$

so conclude that it must be the case that $5^0=1$.

Then we can consider negative exponents, etc.

Answer 3

We have the power law, $$(a^m)^n = a^{mn} \tag 1$$ Furthermore, we have $$(\sqrt{a})^2 = a \tag 2$$ So we can now ask: Does it make sense to write $\sqrt{a}$ as a power, too, and if so, what power does it have to be?

Well, let's answer the second question first, by formally writing $\sqrt{a} = a^x$. Inserting into (2) then gives $$(a^x)^2 = a$$ Now we can apply (1), to get $$a^{2x} = a$$ Now we know $a^1=a$, therefore for the exponential to make sense, we should have $2x=1$, that is $x=\frac12$. Therefore we get $$\sqrt{a} = a^{1/2}$$

Indeed, it is possible to extend the power this way to all rational numbers, that is numbers that can be written as $q=\frac{m}{n}$, by $$a^q = a^{m/n} = \sqrt[n]{a^m}$$

Now to the second question, does it make sense? Well, one can verify that all the power laws still hold, as long as $a>0$, and in those cases one also does not get any contradiction or undefined values. Therefore this definition is, indeed, meaningful.

Answer 4

When we have an initial definition such as "$b^n$ means be multiplied by itself $n$ times" that is only a base definition for whole numbers $n$.

And analogy would be the definition of "$n \times m$ means $n$ added to itself $m$ times". This may be true but then how can we explain "$4/5 \times 1/2$". It is, after all, impossible to add something to itself 1/2 times.

Instead we have the concept of "extending" a definition. If $m \times n$ is $m$ added to itself we can imagine defining a number $v/m$ as "the number that when added to itself $m$ times is equal to $v$" Or in other words "$v/m$ is the number such that $v/m \times m = v$".

But now we have a problem that "$n \times v/m$ means $n$ added to itself $v/m$ times" no longer makes sense.

So we "extend" the definition. We note if $v/m \times m = v$ then $n \times v/m \times m = n \times v$ so "$n \times v/m$ means the number that when added to itself is $m$ times is $n \times v$ or $n$ added to itself $v$ times".

So $b^n$ is the exact same type of situation.

We note $b^nb^m = b^{n+m}$. Now we can't multiply $b$ times itself a zero number of times so $b^0$ by our initial definition makes no sense. But if $b^0$ did equal anything we would have $b^0b^n = b^{n+0} = b^n$ so $b^0 = 0$. We "extend" the definition to include that.

Similarly if we extend the definition to include $b^{-n}b^{n}= b^{n-n} = b^0 =1$ to include "$b^{-n} = 1/b^n$ if $-n < 0$".

If we actually had to spell out what $b^n$ is defined as now it'd be a bit complicated but acceptable: "$b^n$ is $b$ multiplied by itself if $n$ is a positive whole number at least two but if not then if the is another whole number $m$ at least two, so that $n + m$ is a positive whole number at least to then $b^n$ is the number so that $b^{n}b^m = b^{n+m}$ which is just another way of saying $b^1 = b$ and $b^0 = 1$ and $b^{-n} = 1/b^n$ if $-n < 0$ but $n$ is an integer".

Now for $b > 1$ we now that there exists a unique number $\sqrt[n]{b}$ such that $\sqrt[n]{b}^n = b$. We also know that for $n \in \mathbb Z$ that $(b^n)^m = b^{nm}$. We can extend the definition that "$b^{1/n}$ is the number such that $(b^{1/n})^n= b^{n/n} = b^1 = b$... in other words $b^{1/n} = \sqrt[n]{b}$".

This is an extended definition. We no longer are saying "$b^{1/2}$ is $b$ multiplied by itself 1/2 times". We are saying "$b^{1/2}$ is the number that when multiplied by itself $2$ times is $b$". And that is a perfectly valid extended definition.

Just as $"n \times 1/2$ is the number that when added to itself $2$ times is n" is the extended definition when "$n$ added to itself $1/2$ times" no longer makes sense.