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 Extrazlove February 9th, 2019 02:53 AM

new type of number

Hello everyone,

In admit the existence of a negative square thus the existence of i, and that X <= Y and Y> = X without X = Y For the numbers on real.

So

Let the number X such that X <1 and X> 1 with X # 1

And the number Z such that Z <-1 and Z> -1 with Z # -1

The two numbers X and Z are found in the real set with this definition as the case of the number i.

Do X and Z exist as i?

What can we say about X and Z?

 v8archie February 9th, 2019 03:14 AM

We can say they definitely aren't in the reals. What's this new operation \$#\$?

 skipjack February 9th, 2019 04:06 AM

For reals x and y, x \$\small\leqslant\$ y and y \$\small\geqslant\$ x have precisely the same meaning: either x < y or x = y.

For numbers that are imaginary, not real, greater than and less than are undefined.

 Extrazlove February 9th, 2019 04:07 AM

X # 1 means X is different from 1

 skipjack February 9th, 2019 04:14 AM

It's impossible for any real to be both less than 1 and greater than 1 simultaneously.

It's impossible for any real to be both less than -1 and greater than -1 simultaneously.

 Extrazlove February 9th, 2019 04:26 AM

X and Z are not real or complex numbers.
it's a new type of number.

 skipjack February 9th, 2019 04:54 AM

In that case, it doesn't make sense to say "X and Z are found in the real set".

 Extrazlove February 9th, 2019 05:16 AM

The definition of X and Z show that X and Z do not belong to R,
and the use of <and> shows that X and Z do not belong to C either.
Therefore the numbers X and Z are not real or complex.

 skipjack February 9th, 2019 06:46 AM

Can you clarify the definition, as it isn't clear enough for me to understand it?

Whatever definition you use, it doesn't make sense to say X and Z are found in the real set if they're not real.

 Extrazlove February 9th, 2019 06:57 AM

My definition is simple X is greater and less than 1 without being equal to 1.
So X do not belong to R,
and since I used <and> to define it, it does not belong to C either.

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