is .9 repeating and 1 the same size?

15 Answers

Relevance
  • 9 months ago

    If n = .999...

    100n = 99.999....

    100n - n = 99

    99n = 99

    n = 1

    Therefore 1 = .999999999999.........

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    Yes. I can prove this.

    Let n = .99999...

    Let 10n = 9.99999...

    10n = 9.99999....

    n = .99999....

    9n = 9

    n = 1

    • Richard9 months agoReport

      so if the universe is infinitely large, we still have 1 universe; not .999... of a universe

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    They have exactly the same value. They are both equal to the integer 1.

    People conceptually seem to have a problem seeing this wanting to say,

    - "it is *almost* the same", or

    - "it is just less than 1 by 0.0000...0001", or

    - "it *approaches* the same value but never reaches it".

    But those are all people that misunderstand a repeating decimal representation or can't wrap their head around there being a second representation that is equivalent to the whole number 1.

    Take the example of 1/3. If you use long division, you'll get 0.3333... You can't ever finish because it will just continue having a repeated digit of 3. That string of 3s will go on forever. Mathematically 0.3333... (repeating forever) is exactly equal to 1/3. It's not *almost* 1/3. It's not *approaching* 1/3. It *is* 1/3.

    So now look at this:

    1/3 + 1/3 + 1/3 = 0.3333.... + 0.3333.... + 0.3333...

    1 = 0.9999...

    It's pretty obvious this way that they are the same thing. 0.9999... (repeating) *is* exactly the same value as 1. We don't usually write it that way, but technically we could.

    Here's another pattern you can look at. Look at the representations of 1/9, 2/9, etc.

    0.1111... = 1/9

    0.2222... = 2/9

    0.3333... = 3/9 (or 1/3)

    0.4444... = 4/9

    0.5555... = 5/9

    0.6666... = 6/9 (or 2/3)

    0.7777... = 7/9

    0.8888... = 8/9

    0.9999... = 9/9 (or 1)

    Here's yet another way to show it:

    x = 0.99999...

    10x = 9.99999...

    Subtract the two:

    9x = 9

    x = 9/9

    x = 1

    Hence 0.99999... = 1

    Answer:

    They are exactly the same. 0.999... (repeating) = 1; it's not *almost* or *approaching* the same value. It *is* the same!

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    What if x = 0.99999999999999999999999...

    Then 10x = 9.99999999999999999999999...

    Subtract the first equation from the second one:

    9x = 9

    x = 9/9

    x = 1

    So yes, 0.99999999999999999999... = 1

    • Commenter avatarLog in to reply to the answers
  • What do you think of the answers? You can sign in to give your opinion on the answer.
  • 9 months ago

    No. they differ by 0.000....00001

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    yes

    .................................

    • Commenter avatarLog in to reply to the answers
  • David
    Lv 7
    9 months ago

    Only if you round it up to 1

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    Yes. Any rational number can be expressed with trailing zeros or as the number with the last digit in the decimal expansion before those zeros minus one followed by trailing nines.

    • Commenter avatarLog in to reply to the answers
  • david
    Lv 7
    9 months ago

    YES they are the same

    //// think. 1/3 = 0.3333333333... repeating

    multiply both sides by 3

    3x(1/3) = 3x(0.3333333...)

    1 = 0.999999999... Yes they are the same

    • Commenter avatarLog in to reply to the answers
  • 9 months ago

    Not unless you are rounding up. Otherwise, they will never be the same no matter how many 9s come after.

    • ...Show all comments
    • Puzzling
      Lv 7
      9 months agoReport

      There are an *infinite* amount, so yes, they are the same. There is no rounding required.

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.