Comment on I just cited myself.

• ytg@sopuli.xyz ⁨4⁩ ⁨months⁩ ago

  For any a, b, c, if a = b and b = c, then a = c, right? The transitive property of equality.
  For any a, b, x, if a = b, then x + a = x + b. The substitution property.
  By combining both of these properties, for any a, b, x, y, if a = b and y = b + x, it follows that b + x = a + x and y = a + x.

  In our example, a is x’ (notice the ') and b is 0.999… (by definition). y is 10x’ and x is 9. Let’s fill in the values.

  If x’ = 0.9999… (true by definition) and 10x = 0.999… + 9 (true by algebraic manipulation), then 0.999… + 9 = x’ + 9 and 10x’ = x’ + 9.

  if you are rearranging algebra you have to do the exact same thing on both sides

  If you actually change any of the sides. Since, after substitution, the numeric value doesn’t change (literally the definition of equality), I don’t have to do anything – as I’m not rearranging. I’m merely presenting the same value in an equivalent manner. By contrast, when multiplying both sides by 10, since multiplication by 10 changes the concrete numeric value, I have to do it on both sides to maintain the equality relation (ditto for subtracting x’). But substitution never changes a numeric value – only rearranges what we already know.

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  • SmartmanApps@programming.dev ⁨4⁩ ⁨months⁩ ago

    10x = 0.999… + 9 (true by algebraic manipulation)

    No, you haven’t shown that, because you haven’t shown yet that 9x=9. Welcome to why this doesn’t prove anything. You’re presuming your result, then using it to “prove” your result.

    What we know is that the right hand side is 10 times 0.9999…, so if you want to substitute x=0.99999… into the right hand side, then the right hand side becomes 10x (or 9x+x)… which only shows what we already know - 10x=10x. Welcome to the circularity of what you’re trying to achieve. You can’t use something you haven’t yet proven, to prove something you haven’t yet proven.

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