Is the statement |1/0| > 3 valid or invalid?

[uart]

Another annoying one divide zero thread. :)

The statement : | 1/0 | > 3 is,

a) true
b) false
c) invalid



I'm happy enough to say it's true but what do you think?

 

[Njorl]

Sure. Just because something is undefined does not mean we can't make some determinations as to it's nature.

Njorl

 

[matt grime]

1/0 is not a member of the ordered field of real numbers, so which ordered set is this a statement about?

 

[Gokul43201]

Sure. Just because something is undefined does not mean we can't make some determinations as to it's nature.

Njorl

If it's not defined in the reals, it does not exist among the reals.

Alternatively, you can not make determinations of its nature, if its nature is not defined.

...I think.

 

[e(ho0n3]

I think this all depends how you are defining >. If you are defining the relation > on the set of reals, then the statement is invalid since 1/0 is not a real number.

 

[NSX]

What type of number is 1/0 then?

 

[ExecNight]

Invalid..


|1/0| > 3 = ...

1/0 > 3,
1/0 > -3

Looks like true though its invalid because..

╚> Nevermind..Just try to divide your cake with 0 next time before you eat it..And you will see why its invalid..

 

[PRodQuanta]

I don't think there is a category for the number. I would venture to say imaginary, but I don't know how that would work!

Paden Roder

 

[e(ho0n3]

What type of number is 1/0 then?

It isn't. Its undefined.