Is 0.9999 ... = 1?

@Muhammad Ahmed – I agree with Muhammed Ahmed, this is an approximation that approaches 1, but that doesn't mean it ever actually reaches 1. It may get infinitesimally close, but unless it can reach 1, it is not 1; it's approximately 1 "means 0.99999 ≈ 1 . this is true ."

@Muhammad Ahmed – Yes, you are correct that it tends to 1 but not equal. It seems a practical solution for those who believe it equals one would be write out 9.9999...until they reach 1 then reply back to let us know they found 1.

@Muhammad Ahmed – Im Looking for this!

@Muhammad Ahmed – No it equals one.

@Pruthvi Patel – It is .0001 short of "1".Write a comment or ask a question...

@Eddie Spollin – I agree with the above statement if it equals to 1 then why is it a completely different number? Infinite or not logic tells me that 0.999999… is not the same as 1 as you can see although in retrospect it still tends to be the same but still different example 1+2=3 whilst 0.9999999…+2=2.999999… and again this new number is similar actually almost the same but to say that it is exactly the same is undoubtedly false . Now try multiplying them and look at what you get 1 2=2 but 0.999999... 2=1.9999999… so it is common for most people to say it is the same simply because it is easier and we are a naturally lazy species but say the numbers are the same goes against the value they both have there is nothing more to be said I hope you all understand this as I have tried to make it as simple as possible but still there are techinicalities which prevale against ;)

@Kenneth Drew – $0.9999999\neq 1$

It is $\frac { { 10 }^{ x }-1 }{ { 10 }^{ x } }$ that tends to $1$ as $x$ approaches $\infty$ , but because $x$ never reaches $\infty$ , $\frac { { 10 }^{ x }-1 }{ { 10 }^{ x } }$ will always be less than $1$ .

$\frac { { 10 }^{ x }-1 }{ { 10 }^{ x } } =0.9999999$ at $x=7$ .

However, $0.9999999$ is not $0.\overline { 9 }$

$0.\overline { 9 } \neq \frac { { 10 }^{ x }-1 }{ { 10 }^{ x } }$ because $x$ never reaches $\infty$ .

$\frac { { 10 }^{ x }-1 }{ { 10 }^{ x } }$ tends to $0.\overline { 9 }$ as $x$ approaches $\infty$ . It does not become $0.\overline { 9 }$ . It only tends to it.

In other words, $\lim _{ x\rightarrow \infty }{ \left( \frac { { 10 }^{ x }-1 }{ { 10 }^{ x } } \right) } =0.\overline { 9 }$ , but we know that $\lim _{ x\rightarrow \infty }{ \left( \frac { { 10 }^{ x }-1 }{ { 10 }^{ x } } \right) }$ is also equal to $1$ , because $\frac { { 10 }^{ x }-1 }{ { 10 }^{ x } }$ tends to $1$ as $x$ approaches $\infty$ .

$0.\overline { 9 }$ has a repetend, and every number that can be represented with a repetend is a rational number. The repetend of $0.9999999$ is $\overline { 0 }$ , while the repetend of $0.\overline { 9 }$ is $\overline { 9 }$

The rational number that $0.\overline { 9 }$ represents is $1$ . It just so happened that the number $1$ can also be represented as $0.\overline { 9 }$ , though this representation is less popular because it's less simpler to write.

I may have been the only one to notice this, but there is a reason why $1$ can be represented in two ways in Hindu–Arabic numeral system. In decimal notation, if all the digits to the left of the decimal point are $0$ , the absolute value of the real number being represented doesn't get less than the absolute value of it raised to a positive integer. On the other hand, if not all of the digits to the left of the decimal point are $0$ , the absolute value of the real number being represented doesn't get greater than the absolute value of it raised to a positive integer. $1$ neither gets less nor greater than $1$ raised to a positive integer, which is why it can be represented either way. This is also true even in binary, octal, etc. Just an observation on mathematical notation.

$1-0.9999999=0.0000001\\ 1-0.\overline { 9 } =0.\overline { 0 } = 0$

$0.9999999<1$ , but $0.\overline { 9 } =1$ .

@Kenneth Drew – It's not 0.999999999 though. It's 0.999... The trailing dots are important. It means an infinite number of 9s.

@Abhishek Dass – https://www.khanacademy.org/math/recreational-math/vi-hart/infinity/v/9-999-reasons-that-999-1

See this

@Abhishek Dass – yeah, and using your heart, you fail at quantum mechanics. congrats.

btw, in physics, this kind of problem does not even appear, since physics is more about practicality, and hardly has anything to deal with "the exact value". For instance, $\pi$ to the first $30$ decimal places is far from enough to calculate the size of the observable universe to an extremely precise measure.

@Abhishek Dass – I see no further meaning in trying to explain to you bro.

@Abhishek Dass – .99999... simply equals 1. His proof was correct.

@Abhishek Dass – In physics, we look at the problem, conceptualize, solve, and then check the internet to see if we screwed up before handing it in. Do that last step and you will find that all qualified people who internet will disagree with you.

@Abhishek Dass – I agree with Abhishek. I f 0.99999999...... would be EQUAL to 1, then we would always write it as 1 itself. We just round it off and take it to be 1 for convenience. As Karthik said 'The fact that 0.99999....... = 1 is an important result used in Bayes's Calculus too, and hence you cannot say that this result is a contradiction to work on Calculus.' is just because you can't keep on calculating with an infinite series. So we CONSIDER it to be 1. I hope you would agree that 0.9999....... is less than 1. Now, a number less than another can't be equal to it, right?

@Ayran Michelin – Exactly! Math is abstract. It does not necessarily represents only what we can understand/represent in a physical way. Even the concept of infinity already messes with the mind of fresh undergraduates. But, for me, 0.9999... is clearly = 1.

@Karl Walker – A 0.999999... KG weight IS a 1 KG weight.