Fluffy
A fool
Edit: To avoid confusion please note that I am not talking about "0." with a large number of 9's following it but an infinite amount of 9's following it.
Out of curiosity and because of a poll on another forum with the same question, do you believe that 0.9 recurring (that is to say 0.9 with an infinite number of 9s after 0.9) is exactly equal to 1?
The result on the other forum surprised me so I wish to see if it was an anomaly.
The two most common approaches for proving 0.9^=1 ("^" indicates an infinitely recurring decimal)
Out of curiosity and because of a poll on another forum with the same question, do you believe that 0.9 recurring (that is to say 0.9 with an infinite number of 9s after 0.9) is exactly equal to 1?
The result on the other forum surprised me so I wish to see if it was an anomaly.
The two most common approaches for proving 0.9^=1 ("^" indicates an infinitely recurring decimal)
mathematical proof 1 said:1/3 = 0.3^
3 X 1/3 = 3 X 0.3^
3/3 = 0.9^
3/3 = 1
Therefore 0.9^ = 1
mathematical proof 2 said:x = 0.9^ (Stating the definition of x)
100x = 99.9^ (Multiplying both sides by 100)
100x - x = 99.9^ - 0.9^ (Subtracting x from both sides)
99x = 99 (Simplifying the previous step)
x = 1 (Dividing both sides by 99)
Therefore, 1 = 0.9^ (Inserting the value of x back into the original definition)
mathematical proof 3 said:Infinite Convergent Geometric Series: a/(1-r) = a + ar + ar*2 + ar*3...
Let a = 0.9 and r = 0.1.
Therefore, first term = 0.9, second term = 0.09, third term = 0.009. Extended into infinity and totalled, the series equals 0.9^.
Therefore a/(1-r) = 0.9^
Input a and r into left hand side
0.9/(1 - 0.1) = 0.9^
1 = 0.9^