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Feb 12, 2016 at 16:46 comment added lisyarus @ThomasAndrews sounds like your comment should be the answer.
Feb 12, 2016 at 16:45 history closed John B
3SAT
Kamil Jarosz
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Duplicate of What are imaginary numbers?
Feb 12, 2016 at 16:44 comment added N. S. @user51189 no, $i$ is not an integer. It is a complex number which is neither real, nor integer nor rational.
Feb 12, 2016 at 16:43 comment added Gyro Gearloose Integers are defined to be real, but there's an extension: en.wikipedia.org/wiki/Gaussian_integer
Feb 12, 2016 at 16:41 answer added trying timeline score: -3
Feb 12, 2016 at 16:40 comment added Gyro Gearloose Look at en.wikipedia.org/wiki/Field_extension for better understanding. $\mathbb C$ is isomorphic to $\mathbb R[X]/(X^2+1)$. (And be warned, this is a big thing, don't be frustrated if you don't getunderstanding in 10 minutes).
S Feb 12, 2016 at 16:40 history suggested Bobson Dugnutt CC BY-SA 3.0
Grammar and spelling.
Feb 12, 2016 at 16:40 review Suggested edits
S Feb 12, 2016 at 16:40
Feb 12, 2016 at 16:39 comment added Thomas Andrews No, it is not an integer, nor a real number. It is a "complex number." It is something called an "algebraic integer," but that is not the same as being an integer. $\sqrt{2}$ is an algebraic integer, for example, but it is not an integer. Similarly, there is a class of numbers called "Gaussian integers," and $i$ is a Gaussian integer, but again, not an integer.
Feb 12, 2016 at 16:38 comment added zeraoulia rafik if it is a number , is it integer ?
Feb 12, 2016 at 16:36 comment added N. S. $i$ is just a number and should be treated as any other numbers.
Feb 12, 2016 at 16:35 history edited Yai0Phah
edited tags
Feb 12, 2016 at 16:34 history asked zeraoulia rafik CC BY-SA 3.0