Question about the meaning of closed (ops on sets)

[Peppermint]

Hello. I am trying to do a proof and I am using the assumption that addition and multiplication are closed on the set of Real Numbers.

I know that this means if x and y are Real, then x + y is Real.

Can this also mean that if x is not real and y is (maybe) real, that x + y is not Real... because x is not real?

I realized this might be a problem in multiplication. If x is not real and y is not real, I want to say that x * y is not real. But this doesn't seem to be true if x = y = i. Then we have i^2 which is 1, and real.

So am I not thinking of closure correctly?

 

[Mrspi]

Hello. I am trying to do a proof and I am using the assumption that addition and multiplication are closed on the set of Real Numbers.

I know that this means if x and y are Real, then x + y is Real.

Can this also mean that if x is not real and y is (maybe) real, that x + y is not Real... because x is not real?

I realized this might be a problem in multiplication. If x is not real and y is not real, I want to say that x * y is not real. But this doesn't seem to be true if x = y = i. Then we have i^2 which is 1, and real.

So am I not thinking of closure correctly?

If you're dealing with closure, you only want to consider situations where "x" is a member of the specific set, and "y" is also a member of the specific set...and THEN you look at whether x "operation" y is or is not a member of the specific set.

So...if you want to know whether "multiplication" is a closed operation on the set of real numbers, you only consider situations where you start out with two elements of the set of real numbers, and determine whether the product of those two real numbers is a real number.


What happens if you deal with one or two elements which DO NOT BELONG to the set in question doesn't tell you anything about closure.

In other words, if you are trying to decide if a specific operation is "closed" on the set of REAL numbers, then you need only look at situations dealing with two REAL numbers.

 

[Peppermint]

Thank you.

Then I suppose I don't mean closure.

I want to still use the following implication.

if x is not real then x + y is not real.


So I'm saying if x not in R, then x + (anything) is never going to be in R. I really can't think of a way to do a proof on this or any counterexamples.

if x = i, then i + y is in the Complex set, not in R.

(And I don't know any numbers other than C that are not in R.)

So is the implication true or is it unable to be proven true?

 

[Mrspi]

Thank you.

Then I suppose I don't mean closure.

I want to still use the following implication.

if x is not real then x + y is not real.


So I'm saying if x not in R, then x + (anything) is never going to be in R. I really can't think of a way to do a proof on this or any counterexamples.

if x = i, then i + y is in the Complex set, not in R.

(And I don't know any numbers other than C that are not in R.)


So is the implication true or is it unable to be proven true?

If x is a complex number and x IS NOT an element of the reals, then x = a + bi where b is not equal to 0.


Pick another complex number y=c + di.

Add this number to x...
x = a + bi (where b is not equal to 0)

x + y = (a + bi) + (c + di)

x + y = (a + c) + (bi + di)

x + y = (a + c) + (b + d)i

Is it possible that x + y IS an element of the real numbers?

I think it is possible...

(a + c) + (b + d)i will be a real number if (b + d) = 0...or, if b = -d.

Here's an example. Suppose x = 2 + 3i. Then x is clearly not real.

Suppose y = 5 - 3i

Then x + y = (2 + 3i) + (5 - 3i)
x + y = (2 + 5) + (3i - 3i)

x + y = 7

Now...neither x nor y was a real number, but the sum (x + y) IS a real number.