My Math Forum Square roots of numbers < 1

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 March 26th, 2014, 11:18 PM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 429 Thanks: 18 Square roots of numbers < 1 Question) On your calculator, enter a positive number less than 1. Repeatedly press the square-root key. The displayed number should be increasing. Will they ever reach 1? I tried this and the square root increases with each step. It reaches 0.9xyz... very fast (with 0.1 it takes 4 steps). After that, the 10ths digit stays 9 and the 100ths digit approaches 9. This pattern repeats, with each step taking the digits upto but not beyond 9. I imagine that after some n steps the calculator will display 0.999999abcde... This number seems to look like 0.999999...., which is equal to 1. But this number 0.99999.... contains the idea of infinity. No amount of finite steps can equal it. SO Should I conclude that this particular procedure will result in a number that approaches 1 but never equals 1? OR Should I conclude that the procedure will result in a number that will eventually equal 1? ALSO I notice that the procedure looks somewhat like a geometric progression in reverse. Let the number = x such that 0
 March 27th, 2014, 12:20 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 I tried this with the MS Windows calc program. After 109 steps, the result reached 1.
March 27th, 2014, 01:17 AM   #3
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Re: Square roots of numbers < 1

Quote:
 Originally Posted by shunya Question) On your calculator, enter a positive number less than 1. Repeatedly press the square-root key. The displayed number should be increasing. Will they ever reach 1?..................This number seems to look like 0.999999...., which is equal to 1. But this number 0.99999.... contains the idea of infinity. No amount of finite steps can equal it. SO Should I conclude that this particular procedure will result in a number that approaches 1 but never equals 1? OR Should I conclude that the procedure will result in a number that will eventually equal 1?
You can think about the endless amount of digits in 0.999999........ as a process "evolving" or "living in time" in which case it always stays less than 1, or as an eternal or timeless or static infinite amount of nines in which case it is equal to 1. It is the same as with $\pi$, does it exist somewhere in a Platonic world as complete or do we create its digits by calculating it, in which case we can never reach its complete value, it cannot be written down, in which case we may even reach a conclusion that $\pi < \pi$ because we can never reach the true $\pi$
since we have already reached it before we try to reach it.

Also try to think about 1/3=0.33333333333333................ as a process. Can you make an object disappear by dividing it into smaller and smaller
pieces? 1/3 seem to be a complete so nothing is missing from it so that 3*1/3 = 1 and you can get a whole number.
But 3*0.3333333333...........=0.999999999999.......... ....... and there seem to be missing a piece, an infinitesimal.

Now, consider
0.333333....................333......... + 0.333333...................333............ = 0.666666.......................666............
this is true if you think these decimal numbers as "static" or "not evolving in time"
next, think
0.333333...................333......... + 0.333333....................333............ = 0.666666.......................333.............
this is true if the "tails" of the decimals are "living" or are not synchronized to each others, the addition of the digits does not happen
simultaneously at all places.

viewtopic.php?f=27&t=46382

 March 27th, 2014, 02:40 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 That's too convoluted, long-winded and unclear to make any sense at all.
March 27th, 2014, 03:18 AM   #5
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Re:

Quote:
 Originally Posted by skipjack That's too convoluted, long-winded and unclear to make any sense at all.
I took an idea from Rudy Rucker's book Infinity And The Mind, page 254 he writes that:
"$\Omega$ is an ordinal, which means that $\Omega < \Omega$, since every ordinal is less than $\Omega$.
But no ordinal can be less than itself, since you can never count up to something if you have to count up to it before you can count up to it."

At first sight this sentence may seem to make no sense at all, but the more I think about what he writes the more it starts to make sense.
I began to think that this applies also to $\pi$, explaining why we cannot count up to its exact value. We have already counted up to it.

March 27th, 2014, 03:21 AM   #6
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Re: Square roots of numbers < 1

Quote:
 Originally Posted by TwoTwo we may even reach a conclusion that $\pi < \pi$ because we can never reach the true $\pi$
That's the best yet!

 March 27th, 2014, 04:38 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 Note that "every ordinal is less than $\Omega$" is incorrect.
March 27th, 2014, 05:53 AM   #8
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Quote:
 Originally Posted by shunya Question) On your calculator, enter a positive number less than 1. Repeatedly press the square-root key. The displayed number should be increasing. Will they ever reach 1? I tried this and the square root increases with each step. It reaches 0.9xyz... very fast (with 0.1 it takes 4 steps). After that, the 10ths digit stays 9 and the 100ths digit approaches 9. This pattern repeats, with each step taking the digits upto but not beyond 9. I imagine that after some n steps the calculator will display 0.999999abcde... This number seems to look like 0.999999...., which is equal to 1.
"Seems to look like" is not the same as "is"!

Quote:
 But this number 0.99999.... contains the idea of infinity. No amount of finite steps can equal it. SO Should I conclude that this particular procedure will result in a number that approaches 1 but never equals 1?
I would not phrase it that way. "A number" does not "approach" anything! I would say that this procedure results in a sequence of numbers that approaches 1 but is never equal to it.

Quote:
 OR Should I conclude that the procedure will result in a number that will eventually equal 1? ALSO I notice that the procedure looks somewhat like a geometric progression in reverse. Let the number = x such that 0
No, "the number in the calculator display" will never be 1.

More to the point, no number in the sequence will be 1. Strictly speaking a "number in a calculator display" is not a number at all. Since the caclculator display has only a finite number of digits, it really indicates a range of numbers. If, for example, the calculator display allows 10 digits (and the decimal point) then "1.0000000000" indicates any number from 0.99999999995 to 1.00000000005. The field of "fuzzy arithmetic" has been developed to deal with that.

 March 27th, 2014, 11:03 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 The displayed number doesn't "indicate" or "represent" a range of numbers. It's simply the result of the calculator's procedure for calculating an approximate value of the square root of the number previously displayed. The number calculated might or might not be the exact square root of the previous number displayed, and might or might not be "accurate" to the number of digits the calculator is able to display. The procedure described may well eventually lead to the value 1 being displayed. If that occurs, it is likely that all further use of the Sqrt key will also cause 1 to be displayed.
 March 28th, 2014, 03:13 PM #10 Newbie     Joined: Mar 2014 Posts: 13 Thanks: 6 Even though the calculator may reach 1 after a certain length of time, or after hitting the square root key so many times, remember that the calculator can only display so many digits, the true answer will never actually be 1. As far as the calculator goes, say your calculator can display 10 digits. Of course, everyone knows as you keep taking the sqrt of a number, the curve of the line gets closer and closer and closer to 1, but never touches 1; just like a limit. Now, as long as you hit the sqrt key on the calculator on a 10 digit calculator (or however many digits it is,) and one of those numbers is NOT a 0, it will display an answer slightly higher then 1, i.e. 1.000000001, with the final digit being other then 0. Then if you hit the sqrt key again the answer may be 1.0000000001 (notice I added an extra 0 in there.) But since the calculator can only read 10 digits in this case, there may be too many zeros in the actual answer for the calculator to display the true answer, therefore, the calculator is just going to read "1" since there is too many zeros for the calculator's capacity or capabilities. But always keep in mind that the square root of ANY number will never be 1! Hopefully I haven't confused anybody....

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