Complex number confussion

Question

We know that a complex number is represented by the form $a+ib$, where $a$, $b$ are two real numbers. Now $2=2+i.0$ Then $2$ should be a complex number. So my statement is every real number is a complex number. Is the statement true?

Answer 1

Yes, every real number is a complex number.

Answer 2

We can identify complex numbers of form $a+0i$ with real numbers $a$.

Answer 3

$2$ is complex number. Just as natural numbers are is a subset integers which is a subset of the rationals which is a subset of the reals, the reals are a subset of the complex numbers.

In a comment you asked that the reals have an order relation and the complex don't. That's actually fine. If a set has a property that doesn't mean a superset will. For a trivial example consider the set of even numbers. Every even number is an integer. Now every even number is divisible by $2$. But it is not true that every integer is divisible by $2$. Is that a contradiction? Of course not. Being divisible by $2$ is something every even number has but not every integer has.

You commented: "But the real numbers can be compared to each other. As example 2>1. But the complex numbers can't be compared as they represent a point in the complex plane. So what's it? "

If the Complex numbers are represented as points in a plane, the Real numbers are represented as a subset as the point of a line in that plane. We can compare points on a line by looking at which one is further in one direction than the other. We can't do that for points in a plane. We can say "this point has a greater absolute distance from zero than this point" but we can not compare two points that are the same distance but in different directions.

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Oh..... I see your confusion.

It is not true that you can not compare any complex numbers. It's just that in general you can compare every complex number. You can obviously compare $4 + 0i > 2 + 0i$. But you can not compare $2 + 0i$ to $0 + 6i$.

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Actually to get complicated, more complicated than is probably a good idea. We can order and compare complex numbers. We can say $a + bi > c + di$ if $a > c$ or if $a=c$ and $b> d$. This is called the lexicographic order; we compare first the real part and if equal we then compare the imaginary part (but only if the real parts are equal). BUT this order does not have algebraic properties. If $x > y$ and $z > 0$ that we would no longer be able to say $xz > yz$. Of if $x > 0$ and $y> 0$ that would no longer be able to say that $xy > 0$. And without those properties this order just isn't useful to us.

Answer 4

Yes, every real number is also a complex number. Real numbers, denoted by $\mathbb{R}$, are proper subset of complex numbers (denoted by $\mathbb{C}$). Every real number, say $ x$, can be expressed in the general form of complex numbers, which is $ x + iy$, where x represents the real part whereas y represents the imaginary part of the complex number.

$\mathbb{N} \subset \mathbb{W} \subset \mathbb{Q} \subset \mathbb{R} \subset\mathbb{C}$.

Also, remember that real numbers can be compared to each other (e.g., $5>2$), but in the case of complex numbers, we cannot compare them. Even $3+2i$ and $2+2i$ cannot be compared.

That's a brief introduction about complex numbers.