Can you prove #0.bar9 = 1?#

2 Answers
NickTheTurtle · EZ as pi
Jan 26, 2018

Yes, #0.999999…=1#

But note that the above holds true only if the #9#'s are recurring infinitely (which cannot be well-defined with only basic algebra concepts)

Explanation:

Here is one non-rigorous proof.

Consider #x# such that
#x=0.999999…#

Multiply both sides by #10#:
#10x=9.999999…#

Subtract the first expression from the second expression:
#10x-x=9.999999…-0.999999…#
#9x=9#
#x=1#

Thus, #x=0.999999…=1#.

To make it more rigorous, instead of saying #1=0.999999…# where #9# is recurring, we should say #0.999999…#, where the number of #9#'s tends to #oo#, converges to #1#. However, this would involve the concepts of limits and convergence, something beyond algebra.

GM_Phrostbyte
Jan 26, 2018

Yes, #0.bar9=1#

Explanation:

Here is one of the proofs:

#1/3=0.bar3#

#1/3*color(blue)3=0.bar3*color(blue)3#

#3/3=0.bar9#

#1=0.bar9#

Since all of the algebraic manipulations was done correctly, the statement holds true.

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