How Do Negative Exponents Work in Math?

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Negative exponents can be simplified by moving the base to the opposite side of the fraction, effectively turning the exponent positive. The "imaginary 1" mentioned serves as a placeholder to maintain the integrity of the fraction during this process. For example, 7^(-3) can be rewritten as 1/(7^3), confirming the rule that a negative exponent indicates the reciprocal of the base raised to the positive exponent. This principle is consistent with the exponent rule that states 7^a * 7^b = 7^(a+b), which holds true for all integer values, including zero and negative exponents. Understanding this concept allows for easier manipulation of expressions involving negative exponents in various mathematical contexts.
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Recently in class (11 math) we explored exponents in negatives. My teacher placed a question on the board and it said

7*1/7^-7
( 7 times 1 over 7 to the power of -7 )

She said that in order to get rid of a negative, we must imagine that there is an imaginary "1" multipliying the fraction.

1/7^-7 *1

Then we must "flip" the fraction around to eliminate the negative exponent. so it will be

7*1*7^7/1

I am a type of person to want to to know why and how stuff are done, especially in the subject of math. I want to know the reasoning behind. Unfortunately i couldn't ask her that day, and then the weekend came and the problem still remains.
So now i know how to get ride of the negative exponent, but will someone explain to me where the imaginary"1" came from and will i be able to do this in any equation where a negative exponent is present?
 
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Personally, I have never heard of the "imaginary one" but from how your teacher explained it, I think it makes the whole problem a lot harder than necessary. So here's my explanation: whenever a number is to a negative exponent, all you have to do is move it to the other side of the fraction to get rid of the negative. If the negative exponent is on the denominator, move it to the numerator and vice versa.
Ex. (2^5)/(3^-4) = (2^5)(3^4)
All I did was move 3^-4 from the denominator to the numerator, thus making it's negative exponent positive.
 
Makes sense, but that "1" still exists in the fraction if (2^5)(3^4)/1. But I see how it is sort of irrelevant. Thank you, now what if the number was not in a fraction and was just simply;
(7^-3)
will this be correct? 1/(7^3)
 
7^-3 = 1/7^3, yes.
 
The basic idea behind this is the equation

7^a 7^b = 7^{a+b}​

We know this is true for a and b both positive integers, so we'd like to make it true for other values too.

Put b=0 and the only solution is 7^0 = 1. Then put b=-a to see that

7^{-a} = \frac{7^0}{7^a} = \frac{1}{7^a}​
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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