Is 0.9 Recurring Truly Considered Equal to 1 in Mathematics?

  • Level: High School 
  • Thread starter Thread starter curleymatsuma
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
34 replies · 19K views
wsabol said:
1 is a real whole number. 0.999... is a limit. That limit is equal to 1, not the real decimal number 0.9999...(as close as you can get to infinity without getting there, because the infinite term of the sequence ever happen)...9

No, 0.99999... is a notation for a real number. The real number is defined by a limit.

Limits are real numbers.
 
Physics news on Phys.org
micromass said:
No, 0.99999... is a notation for a real number. The real number is defined by a limit.

Limits are real numbers.

Damn you got me. Ok.
 
There is also the http://en.wikipedia.org/wiki/Infinitesimal" approach.

There is a http://en.wikipedia.org/wiki/Hyperreal_number" [itex]\epsilon[/itex] that is smaller than the smallest real number, so we can define the following: [itex]1 - \epsilon = .999...[/itex].

This implies that [itex]1[/itex] and [itex]1 - \epsilon[/itex] (.999...) are different numbers in the hyperreal numbering system.
 
Last edited by a moderator:
Matt Benesi said:
There is also the http://en.wikipedia.org/wiki/Infinitesimal" approach.

There is a http://en.wikipedia.org/wiki/Hyperreal_number" [itex]\epsilon[/itex] that is smaller than the smallest real number, so we can define the following: [itex]1 - \epsilon = .999...[/itex].

This implies that [itex]1[/itex] and [itex]1 - \epsilon[/itex] (.999...) are different numbers in the hyperreal numbering system.


I fear you have not fully understood hyperreals. In the hyperreals, the definition [itex]1-\varepsilon=0.9999...[/itex] is not made. Furthermore, in the hyperreals, there is no such thing as the smallest real number.
 
Last edited by a moderator:
This topic has come up so often here that we have an FAQ that addresses this concept: [thread]507001[/thread].

Thread closed.