- #1

- 1,222

- 22

[tex]1<i[/tex]

[tex]1>i[/tex]

[tex]i<-i[/tex]

[tex]-1<i[/tex]

[tex]-i<i<-1[/tex]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Char. Limit
- Start date

In summary, the conversation discusses the concept of complex numbers as an ordered field and explains why it does not follow the traditional rules of ordering. It is not possible to write "a < b" for general complex numbers a and b, and defining the order by the modulus would also lead to contradictions. The conversation also outlines the three properties that an ordered field must follow, and shows how these properties are not satisfied by complex numbers.

- #1

- 1,222

- 22

[tex]1<i[/tex]

[tex]1>i[/tex]

[tex]i<-i[/tex]

[tex]-1<i[/tex]

[tex]-i<i<-1[/tex]

Mathematics news on Phys.org

- #2

Science Advisor

Homework Helper

- 42,989

- 975

- #3

- 1,034

- 5

I guess, the order relations need to obey some sensible rules. With complex numbers you would always get contradictions. You could try to define the order by the modulus of the complex number. But then the universal law

[tex]a<b\qquad\Rightarrow\qquad a+c<b+c[/tex]

would not hold for all complex numbers. Therefore you wouldn't be able to use the most basic properties for rearranging equations with variables.

- #4

Science Advisor

Homework Helper

- 42,989

- 975

1) if a< b and c is any third member of the field then a+ c< b+ c.

2) if a< b and 0< c, then ac< bc

3) For any two a, b in the field one and only one must be true:

i) a= b

ii) a< b

iii) b< a.

Suppose the Complex number were an ordered field with order "<". Clearly i is not equal to 0 so by (iii) either i< 0 or 0< i.

If i< 0 then, adding -i to both sides, by (i) 0< -i. So by (ii) we must have i(-i)< o(-i) or [itex]-i^2= 1< 0[/itex]. Since this is not necessarily the "usual" order on real numbers that is not yet a contradiction. However, that implies, as before, that 0< -1 and then, multiplying i< 0 by -1, we have -i< 0, contradicting 0< -i above.

Suppose 0< i. Then 0(i)< i(i)= -1. But then 0(-1)< i(-1) so that 0< -i and, adding i to both sides, i< 0, again a contradiction.

- #5

- 1,490

- 0

The statements 1<i and 1>i are not true because the imaginary unit i is defined as the square root of -1, so it cannot be greater than or less than a real number like 1.

The statement i<-i is also not true because the imaginary unit i and its negative -i have the same absolute value, so they are neither greater than nor less than each other.

The statement -1<i is true because the imaginary unit i is greater than -1 on the complex plane, since it has a positive imaginary component.

The statement -i<i<-1 is also true because -i is less than i, which is less than -1 on the complex plane.

In general, inequalities involving complex numbers are not as straightforward as those involving real numbers because complex numbers have both real and imaginary components. When comparing complex numbers, we need to take into account both the real and imaginary parts.

Real numbers are numbers that can be represented on a number line and include all rational and irrational numbers. Complex numbers, on the other hand, include a real part and an imaginary part represented as a + bi, where a and b are real numbers and i is the imaginary unit.

Inequalities for real numbers are expressed using the greater than (>), less than (<), greater than or equal to (≥), and less than or equal to (≤) symbols. For complex numbers, inequalities are expressed using the modulus or absolute value of the complex number.

The graph of an inequality for real numbers is a shaded region on a number line, with the boundary points representing the values that make the inequality true. For complex numbers, the graph is a shaded region on a complex plane, with the boundary points representing the values that make the inequality true.

To solve an inequality involving both real and complex numbers, you first solve for the real numbers using algebraic techniques. Then, you solve for the imaginary numbers by setting the absolute value of the complex number equal to the value on the other side of the inequality and solving for b.

Inequalities are important in mathematics because they allow us to compare quantities and make statements about their relationship. They are also used in many applications, such as optimization problems in economics and engineering, and in determining the behavior of functions in calculus.

Share:

- Replies
- 7

- Views
- 1K

- Replies
- 13

- Views
- 1K

- Replies
- 3

- Views
- 641

- Replies
- 38

- Views
- 3K

- Replies
- 5

- Views
- 1K

- Replies
- 5

- Views
- 894

- Replies
- 18

- Views
- 1K

- Replies
- 1

- Views
- 569

- Replies
- 1

- Views
- 749

- Replies
- 2

- Views
- 1K