0.999... = 1  Prove Me Wrong
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Voting Style:  Open  Point System:  7 Point  
Started:  7/6/2018  Category:  Education  
Updated:  2 months ago  Status:  Post Voting Period  
Viewed:  279 times  Debate No:  116330 
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Argument:
0.999... = 1 Proofs: x = 0.999... 10x = 9.999... 10x  x = 9.999...  0.999... 9x = 9 x = 1 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... Define x. 0.999... < x < 1 Rules: BertrandsTeapot will NOT be allowed to accept this debate. Please disprove/give sufficient enough evidence for anyone to vote for you, and for me to believe 0.999... DOESN"T equal 1.
1 and 0.999 repeating are not the same number. They are numerically different. This is because numbers are exact. 1 is an exact number used to represent an exact amount. True, you may round 0.9999 repeating up to 1, but when you do that, you are rounding, and not being exact. That is why when you solve 79.99999...+40.00000000012321..., and you want to estimate the answer, you would rewrite it as 80+40. However, that is an estimate. Key word, estimate! That means that in this context, the value of 120, or the sum of the rounded numbers, is an indicator used to help find a relative range of what your true sum will be. So, you round up to 1 when using 0.999 repeating, but that does not mean they represent the same amount. Just like 1.0000000001 will never equal just 1. 

?
What have you proved? Nothing. If you know simple math concepts. Maybe in my second proof. But how did I estimate in my first and third proof? Proof one simply shows that 0.999... = 1. You have not even begun to try to prove proof three wrong at all. Please properly engage/actually give sufficient evidence, or STOP wasting my time.
1 x 0.999...=0.999... 1 x 1=1 Yes, it is true that we use 0.666... repeating to represent 2/3, and 1/3 is represented by 0.333... However, this is simply because a true 1/3 or 2/3 of 1 does not exist, and this is roughly what it would be. Does that make sense? I figured you would see the logic in my last argument, but instead you just acted offended by it. Again, 0.999... is a number very close to 1, but it will never be one. Now, if you multiply a number by 0.999... instead of by 1, you will get a very close number, but it will still be a different number. Of course, the calculator will round it to 1, because it is very close to 1, but it technically is not one. There will theoretically be a difference between the numbers, no matter how small it is. So if you use 0.999... repeating to replace 1, it may work on a calculator, but it technically is still never the same as the number 1. 

?(again)
I hoped you would get the message that I accepted your disproof of my second proof. Once again I say that you haven"t even touched upon my first or third proof. Forget about the second, and start disproving the others.
0.999... The reason you cannot is because, since 0.999... is an infinite number, you cannot find the end to infinity, and therefore find the difference between 0.999... and 1, but it would theoretically be x=0.0000...00001 somewhere, but because 0.999 is infinite, you will never know the exact amount of x. So the difference between 1 and x is infinitely small, but still exists. https://hiizuru.wordpress.com... I read the article, and found that it shows that you can look at it from both ways, and still be right. However, by all technicality, I firmly believe that 0.999, even if some people use it to replace 1, does not equal 1, and never will. There will always be a difference between the two numbers, even if it is infinitely small. 

What 0? You simply move the decimal point one place to the left. Why is this so complicated?
I say that proof is still valid. And what about proof 3? This one trips up even the strongest deniers. https://www.youtube.com...
It really comes down to whether you believe there is a difference between the two numbers, as in, whether it is 0 or 0.000...01. 

What about proof 3?
I appreciate having this debate, but I now agree with you to some extent. 0.999... can replace 1, and the only difference is the infinitesimal 0.0...01, if you believe that even exists. Thanks for having this debate. 
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