Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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SMBoy wrote:

A = ...999

B = ...001

A + B = 1

If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

First off "A + B = Would Equal 1.1" is blatantly false. A + B = 1. ...001, by your notation.

Secondly, that assumes that there is a "last digit". In your notation, the 9's clearly stop. There is a "last nine" in A, and in the same spot in B there is a corresponding 1. The problem is that this contradicts with the very definition of "infinity". If there are an infinite number of 9s, there can be no "last nine". That means that there is no corresponding location in B to put the 1, so B just becomes an infinite string of 0s. And 0.000.... (with an infinite number of 0s) = 0. So if B is 0, and A + B = 1, then A = 1.

Stumpe wrote:

1:

1 X 2 = 2

0.9 X 2 = 1.8

0.99 X 2 = 1.88

0.999 X 2 = 1.888 and so on and so forth, so

0.999... X 2 = 1.888...

because no matter how many 9's you add to it, you will always end up adding that many 8's .

I'd recommend checking your basic math again, before trying to attack a complicated concept like 0.999...

0.99 x 2 = 1.98, not 1.88.

0.999 x 2 = 1.998, not 1.888.

No matter how many 9's you add to it, you're not adding that many 8's, you're adding that many 9's (minus 1). So all it works out to is 1.999....8, and a difference of 0.000...2, which is the same argument as 0.999...9 and a difference of 0.000...1. Which still faces the same problem of that "infinitieth" or "last" digit of an infinitely long string of digits. And because of the definition of infinity, there is NO last digit, so the 'number' 0.000...2 = 0, because you can't have a 2 in the 'last spot', since there's always more 0s still.

4/ Dunno

Thanks for your concern for my love life. I considered finding a girlfriend, but then decided that if I did it wouldn't be fair to my wife and son. Especially since my wife is pregnant with our second child right now. I've heard that you're not supposed to get a girlfriend once you're married with a couple kids... something about "cheating".

;-)

What is this thread even about at this point? Besides pointless bickering, I mean...

Quote: "It was the Demestos firms own TV Commercial that Stated the Quote below!

DEMESTOS KILLS 99.999 PERCENT OF ALL KNOWN GERMS"

99.999 =/= 99.999...

100 - 0.001 = 99.999

That doesn't mean that 99.999... < 1000 though, if that's the "penny" you're trying to use this argument to prove.

I think the confusion is that nobody really knows what the point of this post is.

All you're stating is that something doesn't kill all germs, just most. Okay... so what's the problem? That's probably just a legal thing, since if there turns out that something slips through, they can just claim that it was in the 0.001% that they don't kill and avoid getting sued.

No, I think that apparently I'm the one that'd be going a bit overboard... ;-)

It's a mathematical breakthrough! Numbers less than 100 are less than 100. Genius! ;-)

ppinkins wrote:

Sample question:

An employee drove from Lake Charles to a conference in Port Arthur. A total distance from the round trip was 240 miles. The time required to travel one way to Port Arthur was two hours. Due to heavy traffic during the return trip to Lake Charles, an extra hour was required. How much slower was the employee traveling on the return trip?

Start with figuring the length of each leg. Traffic doesn't change the distance, so if the round trip was 240 miles, he travelled half of that in each direction.

Then find out how fast he was travelling on the way there. Speed is simply distance divided by duration. So if you know how far he travelled to get there, and the question tells you how long it took him, you can calculate that.

To find out how fast he was travelling on the way back is much the same. You just have to do one extra calculation to determine how long it took him to get back (since the question gives you the difference in duration, but not the total new duration). Once you have the new duration, you can calculate the speed returning the same way as above.

Now, just compare the two speeds, and you'll know how much slower he travelled coming home!

Try this link: http://www.bioinfo.rpi.edu/~zukerm/cgi-bin/dq.html

Nifty program somebody wrote up to solve whatever problems you ask it.

Thanks. ;-)

John E. Franklin wrote:

And here's the sentence with all the letters, just in case 'Z' or 'Q' was missing.

Cool, thanks. Yeah, I was missing X, V, Q, J. Of course I knew which 4 symbols represented them, just couldn't match which symbol meant which letter. ;-)

henryzz wrote:

i want solutions

are people basicly saying there r no solutions on the web

Barring the use of billions of men, I'd put forward the solution that the word "square" in this case is misused, and he merely had approximately square rectangles at his use. So even if there were an even number of men in the initial squares (12x12 = 144 per square, for example) the giant "square" had 8785 people in it, who would've been standing in a form approximates 93x93 men. But since I don't think they had helicopters, or the capacity for counting that many people at once, it would've merely seemed to be one monstrous square of people, even if it wasn't perfectly geometrically perfect. ;-)

John E. Franklin wrote:

Also, 4 blocks after the the mistake 'Y'/'A', the color on the bottom is orange, though it might look like red, it's not, it's orange.

Here's some more information that may help someone whose looked at this a lot.

Purple = black + blue + yellow

Green = yellow + blue

Red = blue + black

Orange = yellow + black

Hope I'm not ruining the fun.

Yep, had noticed that it was orange and already figured what all the combo-colours mean. I think my problem with that second-last line was that I had the first word wrong, and it was throwing off my parsing of the rest of the sentence because I couldn't figure out where the noun, adjective, etc was. Now I've got the first 3 words, and will probably have the last as soon as I see the letters again (left them on a different computer), so I think I've got it all solved now. Wahoo, go me. ;-)