# A question about infinity [closed]

Last night I could not sleep because I tried to prove that the first and the last decimal digits of infinity (whether positive or negative) must be $9$. If the biggest number is infinity, it must consist of an infinite succession of the highest valued digit, $9$. Is it possible to prove this conjecture?

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## closed as unclear what you're asking by Did, Pedro Tamaroff, Maisam Hedyelloo, Ayman Hourieh, Andrey RekaloAug 7 '13 at 20:30

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Infinity is not a number with a decimal expansion. –  Tobias Kildetoft Aug 7 '13 at 19:49
MSE: where dreams are crushed. –  Git Gud Aug 7 '13 at 19:54
Why not $F$, as in hexadecimal? –  Karl Kronenfeld Aug 7 '13 at 19:55
Yet another example of misuse of the term "philosophical". –  Did Aug 7 '13 at 19:56
Meanwhile, I'm still wondering what color the empty set is. –  Dave L. Renfro Aug 7 '13 at 20:25

Note that by a similar argument, if $\infty$ had a last digit then for every base $b$, that digit has to be $b-1$. In fact, by similar arguments you can show that every digit must be $b-1$ (otherwise, it's one-before would be "non-maximal").
Hey you stole my argument at the end. You owe me $\$\infty$. (For those who can't tell, I am kidding about both points.) – Karl Kronenfeld Aug 7 '13 at 20:16 Infinity is not an actual number, and so doesn't have any digits. It's an abstract concept that means something without limit. - This is essentially right, but you don't explain in what sense. – dfeuer Aug 7 '13 at 19:58 Ah i understand, so my question should be: is the biggest decimal value a succession of an infinit number of the digit '9'? – Chriss Aug 7 '13 at 19:59 Well, there's no biggest decimal value, because it goes on forever! – Sujaan Kunalan Aug 7 '13 at 20:01 @Chriss your question implicitly assumes there is a "biggest decimal value", which there is not. And there is no real number whose decimal expansion has infinitely many digits -- 9 or otherwise -- before the decimal point. – littleO Aug 7 '13 at 20:07 @Chriss they would be just as infinite as each other – Sujaan Kunalan Aug 7 '13 at 20:10 The decimal expansion of a non-negative real number can be viewed as an infinite series: $$\sum_{k = m}^\infty a_k 10^{-k},$$ where for each$k$,$a_k \in \{0,\dots,9\}$. A negative real number is just the additive inverse of a positive one. Extending this notion to allow the expansion to continue forever to the left as well as the right, you get $$\sum_{k = -\infty}^{\infty} a_k 10^{-k}.$$ In this case, if for each$N\in \Bbb Z$there exists an$n\in \Bbb Z$such that$n<N$and$a_k\ne 0\$, you end up with a series that goes to infinity.