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Does 0.999999.....=1 ?
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Let's imagine an infinite number line. What is the probability that if you select a random integer, it is 0. All integers are equally likely. It is 1 over infinity.
What is the probability of selecting any integer? 1 / infinity per integer times infinity (# integers), which equals 1.
However, if 1 / infinity = 0, then you get 0 times infinity, which equals 0.
1 / infinity =/= 0.
Therefore, 1 - 1 / infinity =/= 1 and 0.9999... does not equal 1.
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10X=9.9999...
10X-X=9.9999....-0.9999....
9X=9
X=1
1=0.9999....
1/3=0.333...
3(1/3)=3(0.333...)
1=0.999...
1/9=0.1111
9/9=0.9999
9/9=3/3=1
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I am confused. First you said ".99999.. is only ever equal to 1 when taking about infinity sequence" But then you said ".999999 is a stretching of the infinity that will never equal one but will be tantalizing close enough to round it to one anyways.". Are you agreeing with me or not?
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Are you saying it is impossible to write 1/3 as a number because 0.333... does not equal 1/3? What do you think 9.999...-0.999... is? Why does 0.999...+9=10*(0.999...) if 0.999... is not one?
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I guess the response to this would simply be that 1/3 might be .333 inf. and somewhere in all of it, there is a 4. Again, based on rounding rules, we are willing to overlook .00000001 to round the number to one, even if it isn't.
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"A 0 with an ifanet zeros behind it is just 0."
0 with infinite zeros behind it is 1 over infinity. But is that zero?
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1/3 = 0.333...
So (1/3)*3 = 0.999... = 1.
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1/3*3 = 0.9999...
0.99999... =/= 1
You can't just round it and say it's a truth.
1/(3/3) = 1/1 but 1/3*3 is only equal to it if you round it away from its true value to a lesser amount of figures.
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1/(3/3) = (1/1)/(3/3)
(1/1)/(3/3) = 1/1 = 1
(1/1)*(3/3) = 1/3 * 3 = 0.999...
Therefore, 1/(3/3) = 1/3*3, and 1 = 0.999... This isn't rounding. I only used addition, multiplication, and division.
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X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....But
9X=9
X=1
1=0.9999....
But what if we did 1-0.999..., wouldn't the answer be 0.0...1? Not 0
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1 is at the end of the the decimal. However because 0.999... is infante so must 0.000.... There is no end to the decimal. If there is no end then there can be no one. Having a infanet amount of zeros with no end to place the one is equal to zero with an ifanet amount of zeros which is zero.
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First, in mathematics we use the concept of limit. The limit is the number to which a given sequence of numbers converges.
For example, the sequence 1, 0.1, 0.01, 0.001, ... converges to 0, that is it gets infinitely close to 0, hence its limit is 0. The sequence 1, 1+1/2, 1+1/2+1/4, 1+1/2+1/4+1/8, ... converges to 2, hence its limit is 2
Strict definition of limit is a number such that the difference between it and the sequence gets smaller than any possible value, starting at some point. Does it get closer to some value than 0.000000000000001? Closer than 0.00000000000000000000000000000001? Closer than...? If the answer to all these questions is "yes", then this value is the limit.
Now, let us come back to the original question: does 0.999... equal 1? To answer this, we must first understand what 0.999... exactly means. By definition, it is nothing more than the limit of the following sequence:
0.9, 0.99, 0.999, 0.9999, ...
What is the limit of that sequence? It is easy to see that it is 1. Indeed, 1 - 0.9 = 0.1; 1 - 0.99 = 0.01; 1 - 0.999 = 0.001 - it is just the same sequence as in the first example above, and it converges to 0. So the difference between 1 and this sequence gets infinitely small, hence the limit of this sequence is 1.
Hence, 0.999... = 1. There is no trick here: 0.999... is just a smart way of writing down a limit of a sequence converging to 1.
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Note, however, that not all repeating decimals can be written as a simple number. For example, 0.444... cannot be written as one short number. However, it can be written as a fraction of 4/9. It is easy to see from the following:
0.444.../0.999... = 4/9
0.999... = 1
0.444.../1 = 4/9
0.444... = 4/9
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For precision, it does not.
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As the fathers of calculus has proved, if something is infinitely small it does not exist. Therefore the difference between 0.999.. and 1 is non-existent.
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1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can't not equal 1 and thus 0.999... equals 1.
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2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1
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3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1
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4. Arithmetic proof:
1-0.999... = 0.000...
Since infinitely small numbers do not exist (Norton and Baldwin, 2018), 0.000... equals 0 and because any number subtracted from an equal value is zero, 0.999... equals 1.
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In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.
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1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can not not equal 1 and thus 0.999... equals 1.
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2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1
Therefore 0.999... equals 1.
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3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1
Therefore 0.999... equals 1.
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4. Arithmetic proof:
1-0.999... = 0.000...
Since any number subtracted from an equal value is zero, 0.999... equals 1.
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In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.
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1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can not not equal 1 and thus 0.999... equals 1.
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2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1
Therefore 0.999... equals 1.
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3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1
Therefore 0.999... equals 1.
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4. Arithmetic proof:
1-0.999... = 0.000...
Since any number subtracted from an equal value is zero, 0.999... equals 1.
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In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.
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