Which one is correct? i (imaginary number) > -1, -1 > i or neither? Why?

In DeVry University Math's answer, he states:

If i² < (-1)²,
likewise i < -1.

But just because a² < b², it does NOT follow that a < b:
(5)² < (-6)² or 25 < 36
From this we obviously cannot state that 5 < -6


Here's an example of trying to compare i to 0

Now if we can compare i to 0, then i is either < 0, > 0, or = 0

----------

Case 1: i = 0

Squaring both sides, we get:
-1 = 0
Not true

----------

Case 2: i < 0

Multiply both sides by i.
Since i < 0, it is negative and therefore we must change direction of inequality:
i * i > 0 * i
i² > 0
-1 > 0
Not true

----------

Case 3: i > 0

Multiply both sides by i.
Since i > 0, it is positive and therefore we do not change direction of inequality:
i * i > 0 * i
i² > 0
-1 > 0
Not true

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So i cannot be compared to 0. Since i is neither positive, nor negative, then how can it be compared to any real number?

What we can do is compare the magnitude of two complex/imaginary/real numbers.

Think of a number line: -- −5 --- −4 --- −3 --- −2 --- −1 --- 0 --- 1 --- 2 --- 3 --- 4 --- 5
Any real number can be plotted on the number line
If we have two real numbers n and m, where m is to the right of n, then n < m

So how do we plot an imaginary or complex number?
We introduce a second axis.
We plot real numbers along horizontal axis (x-axis), imaginary numbers along vertical axis (y-axis) and complex numbers in the 4 quadrants.

Every real/imaginary/complex number can be written as α + β i , where α and β are real numbers
To plot a number, simply use coordinate (α, β)
When α ≠ 0 and β ≠ 0, α + βi is complex        (α, β)
When α ≠ 0 and β = 0, α + βi is real              (α, 0)     on x-axis
When α = 0 and β ≠ 0, α + βi is imaginary     (0, β)     on y-axis
When α = 0 and β = 0, α + βi = 0                  (0, 0)     origin

So how do we compare two numbers on this coordinate grid?
We actually compare their magnitudes, i.e. their distance from origin, using distance formula

So magnitude of α + βi = distance from origin (0,0) = √(α² + β²)
This means that all points on a circle of radius r from origin all have the same magnitude

So when comparing i to -1, we can compare their magnitudes:
i = 0 + 1i . . . . magnitude of i = √(0² + 1²) = √1 = 1
-1 = -1 + 0i . . magnitude of -1 = √((-1)² + 0²) = √1 = 1

So even though we cannot compare i and -1, we can compare their magnitudes
They are the same distance from origin
 

I'm sorry to say this, but i think both the above answers are wrong..

2 complex numbers can NEVER be compared.

so answer should be neither..

The imaginary is defined to be:
i= sq root of -1


Then: i^2= -1


Now, you may think you can do this:


i^2= -1* -1=1
But this doesn't make any sense! You already have two numbers that square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS DO THE i-PART FIRST!

Hi Intellectual!

Interesting question!

You could try this approach:

i ___ -1


Definition: i = √-1


√-1 ___ -1

Square both sides.

(√-1)² ___ (-1)²

(i)² ___ (-1)²

Definition: i² = -1

-1 ___ (-1)²

-1 ___ (-1)(-1)

-1 ___ 1

-1 < 1

If i² < (-1)²,
likewise i < -1.

Final Answer: i < -1 (or -1 > i)

Hope this helps!

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