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# 289

Engineering - Part 1/2 - August 23, 2010
Rating: 4.6/5 (143 votes cast)
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This came off meaner than expected. No worries, Tuesday's comic is in favor of engineers :-)

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## 30 Comments

I know 0.999...=1 , but that equation always saddens me as I failed a course which 1st class was about that.

As an engineer AND mathematician, I still thinks it's hilarious!

Even more fun is talking to (not very mathematical) physicists oder engineers about the dirac delta: It's zero everywhere, but on one point it is infinity, so that the integral of this function is 1 :)...

But it isn't equal to 1, it is approximately equal to 1. We engineers understand the importance of accuracy to the thousandths or even ten-thousandths place. We use approximations only when it is safe and reasonable to do so. Yes, it is unnecessary to defend engineers since we are just that awesome (as Mike already knows), and yes it was a funny joke (especially the note underneath), but I still felt compelled to defend our stance on this issue.

1 = 9/9 = 9 * 0.1111... = 0.9999... => 0.9999... = 1 !

He was talking from an engineer's point of view - If you are not supplied with an infinite amount of decimal points, some of your theories start making trouble in the machinery. Sometimes, for a computer, 1+1=2.0001. Those kinds of errors cannot be ignored. People start complaining and machinery breaks down.

I know what he meant, but the dots are indicating a period here, so 0.999... is always and in any case and universes 1, independent of any computer!

Take note of the ellipses, this means the decimal is repeated, it's not just a pause in his sentence.

1 is provably equal to 0.999..., it is not approximately equal.

1 / 3 = 0.333...
3 * 0.333... = 0.999...
1 = 0.999...
Q.E.D.

It is exactly equal. "0.999..." is a symbol, which is, by definition, "the limit of the infinite series 0.9, 0.99, 0.999 etc.". This is the whole idea of infinite decimal notation - it's a symbol for the limit of an infinite series. In our case, the limit of the series 0.9, 0.99, 0.999 etc. is exactly 1. Therefore, 0.999...=1. Not approximately, precisely. This is definition of the notation.
What karl and Evan said above me is also correct, but it doesn't really explain the issue, since then we have to ask why 1/3=0.333... and why 1/9=0.111...

Yes. Better to prove it this way:

Let x = 0.999...
Then 10x = 9.999...
9x = 10x-x = 9.999... - 0.999 = 9
so 9x = 9 or x = 1
Q.E.D.

Even if you approximate, 0.999... = 1 within the limits of the approximation, since you always round up, given that the next digit is a 9.

Maybe it's due to the old Engineer stereotype: "measure with a micrometer, mark with chalk, cut with an axe."

This is the second time this topic has been discussed and the second time you've beaten me to this proof.

My second favorite proof is that .999... is equal to the infinite series .9 +(.9 x .1)+(.9 x .1^2)... the sum of the infinite series is .9/(1-.1) or .9/.9, so .999...=1.

Wait, I thought it was... (using bold to mark repeating digits, since I can't do overlines)
Let x = 0.99
Then 10x = 9.9
9x = 10x-x =

`  9.90- 0.99------  9.09`

so 9x = 9.09 or x = 1.01
Therefore, it follows that 0.99 = 1.01
`;)`

That gag would have worked better had I not completely screwed up the subtraction. (Moreso than the intentional screwing up of infinite digits, I mean.) 9x = 8.91, so x = 0.99. Oy.

That doesn't work because .999...*10 doesn't equal 9.90, it equals 9.99...

I see it as 1 coming to a costume party, trying to be funny but just pissing of everyone else, making it just that much funnier.

Actually, I strongly disagree with the comic.

An engineer would know that even 0.9999999999 or 0.999999998 would be equal to 1. Depending on the epsilon you would chose when handling floating part numbers with computers ;-)

Not all engineers work with computers though. On paper, a number is exactly what it looks like.

i love you!

... uhm, i obviously meant to say that of course 1=.999...
-.- what are you talking about? what do you mean me being gay?

I'm gonna give Tuesday's comic five stars haha

My favourite proof is as follows. Nobody objects 0.3333...=1/3. Now, multiply both sides by 3 to get 0.999...=1

Yeah I always like to use that one. And then, for extra sanity checks on the part of the audience, go on to demonstrate that if 0.9999... = 1 (and it does), then it must follow that 1.9999... = 2, 2.9999... = 3, ...
heh. That kills 'em.

If 0.999... does not equal one, then there should be a number in between 0.999... and one. Anyone care to tell me what it is?

0.999... is equal to 1 in the same sense that these two sets of real numbers are equal:
[0..1) and [0..1]. All the reals between zero and one except one versus all the reals between zero and one including one.
If you've ever been marked wrong on a test for not including the end point when you should have (or vice versa) you know that the two are not equal.
Someone did some hand-waving ("these are not the definitions you are looking for") when they said that the limit of an infinite series is equal to its endpoint. The SIZE of the sets are the same (infinity and infinity plus one), but they are not the same sets.
The inaccuracy between the two is infinitely small: a limit that approaches zero. But is it zero? No, we wouldn't say "approaches" if it ever reached it.

These intervals have no relationship to the equation (except that one can prove [0,1> is not compact using 1=0.999...) They might be isomorphic in category of sets, but this frame is not good enough to examine this problem. You need to consider these sets as metric or topological spaces. Even then, one would have to construct a functor in such a manner that you can interpret numbers as intervals. But your idea is close to Dedekind cut, and you could also show in that manner that 1=0.999...
I'm not sure that you understand the concept of a limit: a limit of a sequence is a POINT (!!) in topological space such that every open neighborhood of the point contains almost all members of the sequence. Since decimal notation is defined such that the sum of a series represented by the notation is equal to the real number represented, one can easily see that the sequence of partial sums 0.9, 0.99, 0.999... converges to 1, so the corresponding sum of the series is equal to 1, by definition, thus we get desired equality 1=0.999...

0,999... IS equal to one.

/engineering student

Using "surreal numbers", "0,999..." is infinitely many infinitesimals away from "1". Also, limits do not exist in this kind of math. And "∞ + 1" ≠ "∞".
*dodges*

Combat engineers?

What the. An Engineer would conclude 0.999999999 = 1 - Engineeers always round to 3 significant digits.

Am I the only one who noticed this? Under the engineer, its written "engineer", under Mike, "Mathematician" and under the table, "Spiked Math"

x=1;
if(0.9999999!=x && 1.0000000001!=x)
{
cout }

else
{
while(1)
cout }

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