does 1.999999 = 2?

Question

Just wanted to know

Answer 1

Now for a productive answer.


Short answer: no, they are not equal.

Better answer: as mentioned above, if there are an infinite number of 9's, then yes, they are equal, and here's why.

Suppose
x = 1.99999.... (forever)

Then multiply both sides by 10 to get
10x = 19.99999... (forever)

Subtract the top equation from the bottom equation. The left side is 10x - x = 9x. The right side is 19.99999.... - 1.99999..... On the assumption that you can actually subtract infinitely repeating decimals that match and get zero, this difference is 18.

So the new equation you get is
9x = 18.

Divide by 9 to get x = 2. But x = 1.9999.... also. By the transitive property of equality (english translation: if two things equal a third thing, the original two things are equal), you must have 2 = 1.9999....


There's anotehr proof involving infinite series, but this one is more easily accessible to those with limited algebra skills.

Answer 2

Yes, assuming you mean an infinitely repeating series of nines after the decimal place. .99999999999.... = 1.

How's that? They certanly don't look the same. It's an issue of limits and precision. I'll go through some examples.

Suppose you decided to define equality as being within 15% of the exact same value. This sounds silly, but bear with me. By that standard .8 does not equal 1, but .9 does. Obviously, though, this is not true equality.

So let's make the rule ten times more accurate, and insist that numbers be within 1.5% to be considered equal. Now, .98 does not 'equal' 1, but .99 does.

Let's imagine doing this, oh, say 10 times more, so a number has to be within .00000000015% to be considered equal. Now, 0.9999999998 does not equal 1, but 0.99999999999 does.

As you can see, you can increase the precision of your test for equality arbitrarily, and demand any level of accuracy you want, and by adding enough figure 9's after the decimal place you can always get closer to the target. You need 1 part in a million? Write a bunch of nines. One part in a googol (that's 1 with a hundred zeros)? Write 101 nines after the decimal place and you've got it.

Since an infinitely repeating decimal sequence is, er, infinitely repeating, it can always match, no matter what precision you demand.

Here's another way to look at it: Any infinite repetition of one digit after the decimal point is equal to that number divided by nine: .111111111111111111111... = 1/9. .444444444444444... = 4/9 and so on. This also holds true for nine: .999999999999999999... = 9/9; but any number divided by itself is one. So .99999999999999999... = 1.

Answer 3

1.999999 does not = 2
It will never equal 2, however if it is a recurring 9 sequence, then it could be considered being equal to 2. This is because the difference between 1.999999999999999999999 (as it were) and 2 is in essence actually neglible. If it was critically essential then they would not be considered to be equal. So the right answer for your question is actually no.

Answer 4

If the line of 9's stops, then no.
2 - 1.999999 = .000001

However, if the 9's go on forever, then, yes. That is because
2 - 1.999... = 0 because it differs from zero by an amount that can be smaller than any number named. Since the difference can be made arbitrarily small, the difference must be zero.

Answer 5

It does not equal to 2 anymore than 0.999... equals to 1.

If anyone tries to tell you it is true if the nines go on forever, then you know they are lying, because no one can sum an infinite series. We can only find the limits in special cases.

Answer 6

No, 1.999999 does not equal 2.

1.999999 = 1999999 / 1000000, which is very close to 2.

Now, had you asked, does 1.999999... (repeating forever) equal 2, that answer would have been yes, but without the ellipsis (...) following the last digit you wrote, you have a finite number of digits, making it easily convertible into the fraction 1999999 / 1000000.

Answer 7

2 - 1.999999 = 0.000001 So mathematically the answer is no.
Practically yes. Most of the time the equality is good enough.

Th

Answer 8

No, unless it is repeating. There is a definition that explains it, but I forget it so I will try to explain something that helped me believe .99...=1. Okay, we know that 1/3=.33..., also 1/3+1/3+1/3=3/3(or 1). Now try .33...+.33...+.33.... It equals .99..., but we know by the previous equation that it also equals 1, so .99...=1. I hope that helped someone understand better.

Answer 9

Assuming that the 9 repeats infinitely, the proof goes like this.

1.999...*10=19.999...

19.999...-1.999...=18

18/9=2

You can see that 10x-x=9x, so you can put in any number for x and it will still be true. It just so happens that for repeating decimals, the process simplifies them.

You could also just reason that if 3/9=.333... then 9/9=.999...
Since 9/9 is one, the implication is clear.

Answer 10

1.999999 does not equal 2
1.999999... = 2 i.e. repeating 9's, does equal 2.

The difference is that the first is the same as 1.999999000... You didn't indicate that the digit 9 repeats infinitely.