dsgrue3 said:
None of what you said is applicable at all. Not sure why you insist that it is. Given x = 0.999... Show me where y is in this? |
The point is to show you that multiplying, dividing, adding, or subtracting will not give extraneous solutions like squaring will.
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But I'll play along. Do you agree with the following:?
if 10x = 10y, then x = y. ALWAYS
if x - c = y - c, the x = y. ALWAYS
if x/9 = y/9, then x = y. ALWAYS
and vice versa
if x = y, then 10x = 10y. ALWAYS
if x = y, then x - c = y - c. ALWAYS
if x = y, then x/9 = y/9. ALWAYS
Your answer should be yes to all of the above. Now, let's go back to what I did:
x = 0.9999...
if x = y, then 10x = 10y, so we KNOW:
10x = 10(0.9999...) = 9.999...
10x = 9.9999....
if x = y, then we know x - c = y - c. (in this case, c = x = 0.999...), so we KNOW:
10x - x = 9.9999 - x
9x = 9.99999 - 0.9999 (because x = 0.9999)
9x = 9
if x = y, then x/9 = y/9. so we KNOW:
9x/9 = 9/9
x = 1
since x = 0999999 and (not or) x = 1, then 1 = 0.99999...
It has to be true or algebra is fundamentally flawed.
Nothing I did above is wrong. Absolutely nothing.
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You cannot do the same with squaring because if x^2 = y^2, x only = y SOMETIMES.
Fortunately, I did not square. I used operations which were ALWAYS true.
