The negative reciprocal of 2.3 is ___________. *
Answers
Answer:
This simply happens to be a special case in which some number is equal to its negative reciprocal. We can use a little bit of algebra to rearrange things a bit to see what might be going on:
Multiply both sides of the equation by xy, and we get:
-x^2 = y^2
(minus x squared equals y squared)
But how can this be possible? Isn’t “anything” squared always positive (or at least non-negative)? Take any non-zero real number (positive or negative), square it, and you always get a positive number. So what’s going on here?
This equation actually *can* be satisfied if we allow for “imaginary numbers” to be used. For example, the imaginary number (usually referred to as “i”) is the square root of -1. What happens if we set the value of x to be equal to i? Then we would have:
-(i^2) = y^2
-(-1) = y^2
1 = y^2
y = +1 or -1
Similarly, we could set y = i and then solve for x:
-x^2 = i^2
-x^2 = -1
x^2 = 1
x = +1 or -1
BTW, it’s important to note that in this scenario, we have 2 unknowns, but only 1 equation. This means that there are an infinite number of valid solutions to this problem (i.e., an infinite number of values that we could plug in for x and then solve for y, or an infinite number of values that we could plug in for y and then solve for x). Here, we just picked the value of i = sqrt(-1) because it was easy to do so. We could just as easily plugged in 2i or 3i or any other real number times i and done the same thing. What would happen if we had plugged in the value of -i instead of i? We’d get the same results, because (-i)^2 = (-1)^2*i^2 = 1*(-1) = -1. (This is because sqrt(-1) is actually +i or -i.)
To summarize: This problem has much more to do with imaginary numbers being the square root of negative real numbers, than it has to do with “negative reciprocals”. The only thing that the negative reciprocal expression did was obfuscate/mask what is really going on.
NOTE: There are even more solutions to this equation than what I mentioned above. We could actually plug in *any* complex number (a real number plus some imaginary number) into either x or y, then solve for the other variable. Something of the form: x = a + bi (then solve for y), or y = a + bi (then solve for x). But for now, I’ll leave this for a separate discussion because it requires more advanced knowledge of complex numbers: additional manipulations with complex number conjugates, and bisecting the angles of the vector representations of the complex numbers to solve the equations. (A graph of the numbers in the complex plane would also be appropriate to illustrate what’s really going on.)