Not a proof, just wrong. In the “(substitute 0.9999… = x)” step, it was only done to one side, not both (the left side would’ve become 9.99999), therefore wrong.
As I said, they didn’t substitute on both sides, only one, thus breaking the rules around rearranging algebra. Anything you do to one side you have to do to the other.
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.
(Edit)
Take the following simple system of equations.
5y = 3
x + y = 6
How would you solve it? Here’s how I would:
\begin{gather*} %% Ignore the LaTeX boilerplate, just so I could render it
\begin{cases}
y = \frac{3}{5} \\ % Isolate y by dividing both sides by 5
x = 6 - y % Subtract y from both sides
\end{cases} \\
x = 6 - \frac{3}{5} \\ % SUBSTITUTE 3/5 for y
x = 5.4 \\
(x, y) = (5.4, 0.6)
\end{gather*}
My favorite thing about this argument is that not only are you right, but you can prove it with math.
Except it doesn’t. The math is wrong. Do the exact same formula, but use .5555… instead of .9999…
Guess it turns out .5555… is also 1.
Lol you can’t do math apparently, take a logic course sometime
Let x=0.555…
10x=5.555…
10x=5+x
9x=5
x=5/9=0.555…
Oh honey
you have to do this now
Not a proof, just wrong. In the “(substitute 0.9999… = x)” step, it was only done to one side, not both (the left side would’ve become 9.99999), therefore wrong.
They multiplied both sides by 10.
0.9999… times 10 is 9.9999…
X times 10 is 10x.
10x is 9.9999999…
As I said, they didn’t substitute on both sides, only one, thus breaking the rules around rearranging algebra. Anything you do to one side you have to do to the other.
The substitution property of equality is a part of its definition; you can substitute anywhere.
And if you are rearranging algebra you have to do the exact same thing on both sides, always
And if you don’t then you can no longer claim they are still equal.
For any
a
,b
,c
, ifa = b
andb = c
, thena = c
, right? The transitive property of equality.For any
a
,b
,x
, ifa = b
, thenx + a = x + b
. The substitution property.By combining both of these properties, for any
a
,b
,x
,y
, ifa = b
andy = b + x
, it follows thatb + x = a + x
andy = a + x
.In our example,
a
isx'
(notice the'
) andb
is0.999…
(by definition).y
is10x'
andx
is9
. Let’s fill in the values.If
x' = 0.9999…
(true by definition) and10x = 0.999… + 9
(true by algebraic manipulation), then0.999… + 9 = x' + 9
and10x' = x' + 9
.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.(Edit)
Take the following simple system of equations.
How would you solve it? Here’s how I would:
Here’s how Microsoft Math Solver would do it.