Imaginary numbers, as we all know, does not exist. But how can we compare imaginary numbers with real numbers?

Let \( i= \sqrt{-1} \) , let's find out wether if it is positive or negative. \[ |i| = \sqrt{i^2} = \sqrt{-1} =i \] Since \( |i| = i \), therefore \( i>0 \). Since \( i>0, 1>0\), therefore \( i+1>0\), and \( |i+1| = i+1 \) \[ |i+1| = \sqrt{(i+1)^2} = \sqrt{i^2+2i+1} = \sqrt{-1+2i+1} = \sqrt{2i} = i+1 \] Here comes the real thing, find \( |i-1| \) \[ |i-1| = \sqrt{(i-1)^2} = \sqrt{i^2-2i+1} = \sqrt{-1-2i+1} = \sqrt{-2i} = \sqrt{-1} \sqrt{2i} \] Now i know that \( \sqrt{a} \sqrt{b} = \sqrt{ab} \) doesn't always happen, but the equality holds true only except when \( a<0, b<0 \), but since \( 2i>0 \), therefore the equality holds true. Now let's continue. \[ \sqrt{-1} \sqrt{2i} = i(i+1) = i^2+i = i-1\] Since \( |i-1|=i-1\), therefore \(i-1\) is bigger or equal than 0, \(i\) is bigger or equal than \(1\). Since \(i^2=-1\), \(1^2=1\), therefore \(i\) and \(1\) can't be equal, and thus, \( i>1 \).

Am I right?

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TopNewest\(|i| = |0+1*i|=\sqrt{0^2+1^2}=\sqrt{1}=1\)

\(|i+1| = |1+1*i|=\sqrt{1^2+1^2}=\sqrt{2}\)

\(|i-1| = |-1+1*i|=\sqrt{(-1)^2+1^2}=\sqrt{2}\) – Ton De Moree · 3 years, 9 months ago

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What you are trying to show is that there is no way to order the complex numbers in a manner which is consistent with the ordering of the reals, and where the usual rules of how ordering relates to addition and multiplication still hold. You can show this even more easily than you did. If you could find this ordering, then \(i\) would be either positive or negative, in which case \(-1=i^2\) would have to be positive, which is a contradiction.

The complex numbers do exist! It all depends on what you mean by a number. Mathematicians work quite hard to set up what even the real numbers look like. The collection of \(2\times2\) matrices of the form \[ \left( \begin{array}{cc} a & -b \\ b & a \end{array}\right) \qquad \qquad a,b \in \mathbb{R} \] add, subtract and multiply just like complex numbers, so you could identity the matrix above with the complex number \(a+ib\) and have a concrete collection of objects whose algebraic properties were those of the complexes. If you think about what complex modulus and argument do under multiplication, and how to describe enlargements and rotations in \(\mathbb{R}^2\), you will see where this example came from. – Mark Hennings · 3 years, 9 months ago

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Also, look at the third line. How did you got \(-1\)? – Fahad Shihab · 3 years, 9 months ago

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You made a big conceptual error on the third line – Matthew Fan · 3 years, 9 months ago

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– Takeda Shigenori · 3 years, 9 months ago

Which line is the third line?Log in to reply

according to logic, \(i=\sqrt{-1}\) doesn't exist. So, what do we call for 'nothing'? zero, right? So, we can simply say zero for \(\sqrt{-x}\) i am not making conceptual mathematics intervene here, but instead, saying what we experience. – Fahad Shihab · 3 years, 9 months ago

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– Tan Li Xuan · 3 years, 9 months ago

The square root of \( -1 \) is not 'nothing' or zero,it is an imaginary number.And for your other comment,\( i^{2} = -1\) because \( i\) is the square root of \( -1 \).Log in to reply