Why ((+ or -)any numeric value)/0.0 gives (+ or -)infinity? Shouldn't it be an ArithmeticException?

Answer 1

That's one good question.
In Java, for floating point numbers, certain special values have been defined, such as 'NaN' and infinity, for operations which may not give a simple mathematical result.
Hence dividing by 0.0 will give you infinity (it's been defined that way)
However if you try it with 0, i.e. try 1/0, you'll get an arithmetic exception, because such values haven't been defined for integers.

Answer 2

The picking of infinity is not just a random choice. If you divide by smaller and smaller numbers, you see a trend of approaching infinity so making 0 be infinity is not that huge of step.

Answer 3

Weird but true 🤷

Answer 4

John Wells Thanks! I get the gist!

Answer 5

Vedant Bang Thanks. However, could you please tell me what would create a 'NaN' value as you mentioned?

Answer 6

Sadman Sakib Here is a code which ends up in NaN

https://code.sololearn.com/cW6yGYPiiUWp/?ref=app

Answer 7

More details
https://stackoverflow.com/questions/5291606/why-does-1-0-give-error-but-1-0-0-returns-inf

Answer 8

Ḿổḩảḿḿẻď Ấḿiň It didn't really answer my question!

Answer 9

Vedant Bang Thank you! 👍

Answer 10

infinity is just neumaric