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[-] Valthorn@feddit.nu 95 points 1 month ago* (last edited 1 month ago)

x=.9999...

10x=9.9999...

Subtract x from both sides

9x=9

x=1

There it is, folks.

[-] barsoap@lemm.ee 69 points 1 month ago* (last edited 1 month ago)

Somehow I have the feeling that this is not going to convince people who think that 0.9999... /= 1, but only make them madder.

Personally I like to point to the difference, or rather non-difference, between 0.333... and ⅓, then ask them what multiplying each by 3 is.

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[-] Blum0108@lemmy.world 48 points 1 month ago

I was taught that if 0.9999... didn't equal 1 there would have to be a number that exists between the two. Since there isn't, then 0.9999...=1

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[-] Shampiss@sh.itjust.works 23 points 1 month ago

Divide 1 by 3: 1÷3=0.3333...

Multiply the result by 3 reverting the operation: 0.3333... x 3 = 0.9999.... or just 1

0.9999... = 1

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[-] jerkface@lemmy.ca 76 points 1 month ago
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[-] pruwybn@discuss.tchncs.de 68 points 1 month ago

I thought the muscular guys were supposed to be right in these memes.

[-] myslsl@lemmy.world 65 points 1 month ago

He is right. 1 approximates 1 to any accuracy you like.

[-] pruwybn@discuss.tchncs.de 25 points 1 month ago

Is it true to say that two numbers that are equal are also approximately equal?

[-] SpeakerToLampposts@lemmy.world 28 points 1 month ago

I recall an anecdote about a mathematician being asked to clarify precisely what he meant by "a close approximation to three". After thinking for a moment, he replied "any real number other than three".

[-] mpa92643@lemmy.world 23 points 1 month ago

"Approximately equal" is just a superset of "equal" that also includes values "acceptably close" (using whatever definition you set for acceptable).

Unless you say something like:

a ≈ b ∧ a ≠ b

which implies a is close to b but not exactly equal to b, it's safe to presume that a ≈ b includes the possibility that a = b.

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[-] LodeMike@lemmy.today 44 points 1 month ago

0.9<overbar.> is literally equal to 1

[-] UnderpantsWeevil@lemmy.world 19 points 1 month ago

There's a Real Analysis proof for it and everything.

Basically boils down to

  • If 0.(9) != 1 then there must be some value between 0.(9) and 1.
  • We know such a number cannot exist, because for any given discrete value (say 0.999...9) there is a number (0.999...99) that is between that discrete value and 0.(9)
  • Therefore, no value exists between 0.(9) and 1.
  • So 0.(9) = 1
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[-] magic_lobster_party@kbin.run 34 points 1 month ago

If 0.999… < 1, then that must mean there’s an infinite amount of real numbers between 0.999… and 1. Can you name a single one of these?

[-] ns1@feddit.uk 23 points 1 month ago

Sure 0.999...95

Just kidding, the guy on the left is correct.

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[-] rustyfish@lemmy.world 31 points 1 month ago

Remember when US politicians argued about declaring Pi to 3?

Would have been funny seeing the world go boink in about a week.

[-] Wxnzxn@lemmy.ml 25 points 1 month ago

To everyone who might not have heard about that before: It was an attempt to introduce it as a bill in Indiana:

https://en.m.wikipedia.org/w/index.php?title=Indiana_pi_bill

[-] ValiantDust@feddit.de 30 points 1 month ago

the bill's language and topic caused confusion; a member proposed that it be referred to the Finance Committee, but the Speaker accepted another member's recommendation to refer the bill to the Committee on Swamplands, where the bill could "find a deserved grave".

An assemblyman handed him the bill, offering to introduce him to the genius who wrote it. He declined, saying that he already met as many crazy people as he cared to.

I hope medicine in 1897 was up to the treatment of these burns.

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[-] humdrumgentleman@lemmy.world 29 points 1 month ago
[-] HexesofVexes@lemmy.world 14 points 1 month ago* (last edited 1 month ago)

Reals are just point cores of dressed Cauchy sequences of naturals (think of it as a continually constructed set of narrowing intervals "homing in" on the real being constructed). The intervals shrink at the same rate generally.

1!=0.999 iff we can find an n, such that the intervals no longer overlap at that n. This would imply a layer of absolute infinite thinness has to exist, and so we have reached a contradiction as it would have to have a width smaller than every positive real (there is no smallest real >0).

Therefore 0.999...=1.

However, we can argue that 1 is not identity to 0.999... quite easily as they are not the same thing.

This does argue that this only works in an extensional setting (which is the norm for most mathematics).

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[-] someacnt_@lemmy.world 13 points 1 month ago

Are we still doing this 0.999.. thing? Why, is it that attractive?

[-] Leate_Wonceslace@lemmy.dbzer0.com 13 points 1 month ago

People generally find it odd and unintuitive that it's possible to use decimal notation to represent 1 as .9~ and so this particular thing will never go away. When I was in HS I wowed some of my teachers by doing proofs on the subject, and every so often I see it online. This will continue to be an interesting fact for as long as decimal is used as a canonical notation.

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[-] bluewing@lemm.ee 11 points 1 month ago

Meh, close enough.

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this post was submitted on 27 Jun 2024
817 points (95.2% liked)

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