.999...(repeating) =1

Introduction: In order to prove that 0.999...=1, I will be showing this via 3 different mathematical methods. While the numerical portion of this argument is the main argument, I would briefly try to explain using words how exactly .999...=1. 0.999 repeating is a number that can only exist within the realm of mathematics. 0.999 repeating is a number with infinite accuracy, so it's not something that can possibly exist in real life. However, using mathematics, a number that has an infinite number of "9s" after the decimal place would evaluate to 1. This is simply because the limit of 0.999 repeating written as an infinite geometric series would evaulate to 1, something I will show later in this argument. 0.99... may not seem like it ever reaches 1, but if we evaluate the limit of a geometric series that represents 0.999 with an infinite number of 9s, it would indeed be calculated as 1. C.1 Fractions This is the simplest argument to show that 0.999...=1. 1/3=0.333....Adding 1/3=Adding 0.333...2/3=0.666...Adding 1/3=Adding 0.333...3/3=0.999....And as it turns out, 3/3 which simplifies to 1, does indeed equal 0.999.... C.2 Algebra x=0.999...10x=9.999....(Multiply by 10 to both sides) 10x=9+0.999....(We broke up 9.999...to 9+0.9999) 10x=9+x (0.999...is equal to x, as seen in step 1) 9x=9 (Subtract both sides by x)x=1 (Divide both sides by 9) C.3 Limits and geometric series Let's try writing 0.99...as a fraction, and see what we get. Since this number isn't irrational, it can be written as a fraction. In order to do that, we would have to rewrite this number as an infinite geometric series, and then find the sum of that series. The limit of that series as the number of terms tend to infinity, would be 1. While the limit isn't telling you the value of the function at that number, it is telling you what that function is approaching as you get closer and closer to some number (in this case, it's infinity). Since 0.999...as a geometric series has an infinite number of terms, the limit would tell us what the value if 0.999...had an infinite number of "9s". Which it does.Sum of finite geometric series: Sn=a1((1-r^n)/(1-r) Where Sn=Sum of n terms a1=First number in seriesn=Number of terms r=common ratio If con requests, I can derive this formula. However, if we assume that r is greater than 0 and less than 1, we can take the limit of the function as n tends to infinity. Therefore, the sum of an infinite number of terms would be...Sinf.= a1/(1-r). (r^n=0, as n tends to infinity. Therefore, we're left with a1/(1-r))If con requests, I can do a more rigourous proof, though this is the formula for the sum of an infinite geometric series. So, we can now rewrite 0.9999... as...0.9+0.09+0.009+....Which is equal to 9/10+9/100+9/100+....In this series, a1=9/10r=1/10Since |r|<1, we can use the formula of Sinf.Sinf.=9/10/(1-1/10)=9/10/(9/10)=1And that's it. Using the definition of the sum of an infinite geometric series, we figured out that 0.999...as a fraction does indeed evaluate to be 1. Since we can rewrite 0.999 as an infinite geometric series, we can evalaute it's sum in order to figure out it's fractional form. And that form turns out to be 1/1. This debate doesn't really need sources, since it's just math. However, if Con likes, I can show sources of the formula for the sum of a finite and infinite geometric series.