x= 3.999...
10x= 39.999...
10x-x=39.999...-3.999...
9x=36
x=4
so 3.999 = 4
is there a mistake in the equation? therefore does 3.999...=4?
2006-11-27 03:27:53 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
There is no mistake in your answer. This is just one of the wonders of mathematics. Just like 1/3 + 1/3 + 1/3 = 1. But when 1/3 becomes decimal, we write the equation as 0.333... + 0.333...+ 0.333... = 9.999.... Where actually 9.999... = 1. It's same line of reasoning.
2006-11-27 03:34:50 · answer #1 · answered by Sheila 2 · 0⤊ 2⤋
Perfectly valid.
Every real number has an nth decimal for every integer n. x=3.999.... means that the unit digit of x is 3 and every decimal of x is a 9.
Now, suppose x was not quite equal to 4. Say that x<4 somehow. Then there is a number e, which is between x and 4. x
Let's look at the digits of e. Its unit digit must be a 3, since it's smaller than 4.
The 1st decimal of e must be a 9, otherwise it would be smaller than x. What about the nth digit of e? If it were anything less than 9, then we would have e
Therefore x = 3.99999... = 4. It is not "very close" to 4, it does not "tend towards" 4, it IS 4. Show me a number that is any wee bit less than 4 and I will show you that this number cannot be 3.99999... If you show me the number 3 followed by a gazillion trillion to the gazillionth power of nines, I will show you a number that is smaller than 3.9999....
2006-11-27 03:44:17 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Calculus would call 3.999... as a number approaching 4 in a positive direction. When used in another equation it would be true to simply call the number 4. However I would carry a note with the new equation stating how 4 is defined.
2006-11-27 03:35:41 · answer #3 · answered by urbaal_99 2 · 0⤊ 0⤋
There is no mistake. Any number with a repeating 9 on the right of the decimal is equal to the number plus 1 in the place just before the decimal. This same solution method can be used in other repeating decimal cases:
x = .371371371371...
1000x = 371.371371...
------------------------------
999x = 371
x = 371/999
2006-11-27 03:32:15 · answer #4 · answered by Dave 6 · 0⤊ 0⤋
there's no mistake
your proof is a classic
what's key is to understand that as the 9's go "to infinity", it makes no difference if you multiply by 10 (some people seem to believe that your number would have to become 39.999...0, the zero coming "after infinity")
2006-11-27 03:31:08 · answer #5 · answered by AntoineBachmann 5 · 0⤊ 0⤋
10x = 39.99 not 39.999
so 10x - x = 39.99-3.999 = 35.991
9x = 35.991
x = 3.999 not 4
mathematics still reigns accurate
2006-11-27 03:41:11 · answer #6 · answered by Doug R 5 · 0⤊ 2⤋
It's no mistake, though it may take calculus to understand better.
2006-11-27 03:30:07 · answer #7 · answered by Anonymous · 0⤊ 0⤋
you are completely correct
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2006-11-27 03:31:28 · answer #8 · answered by Anonymous · 0⤊ 0⤋