There's a Real Analysis proof for it and everything.

Basically boils down to

• If 0.(9) != 1 then there must be some value between 0.(9) and 1.
• We know such a number cannot exist, because for any given discrete value (say 0.999...9) there is a number (0.999...99) that is between that discrete value and 0.(9)
• Therefore, no value exists between 0.(9) and 1.
• So 0.(9) = 1

He is right. 1 approximates 1 to any accuracy you like.

Computers can calculate infinite series as well as anyone else

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series.

In the case of 0.999..., it can be shown to converge toward 1 with the convergence rule for geometric series.

If |r| < 1, then:

ar + ar² + ar³ + ... = ar / (1 - r)

Thus:

0.999... = 9(1/10) + 9(1/10)² + 9(1/10)³ + ...

= 9(1/10) / (1 - 1/10)

= (9/10) / (9/10)

= 1

Just for fun, let's try 0.424242...

0.424242... = 42(1/100) + 42(1/100)² + 42(1/100)³

= 42(1/100) / (1 - 1/100)

= (42/100) / (99/100)

= 42/99

= 0.424242...

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point.

Somehow I have the feeling that this is not going to convince people who think that 0.9999... /= 1, but only make them madder.

Personally I like to point to the difference, or rather non-difference, between 0.333... and ⅓, then ask them what multiplying each by 3 is.