what is the flaw with this proof?

[soandos]

x==Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]
x==Sqrt[x]
x==1 and zero
1==Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]==0
1==0

[CRGreathouse]

x==Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]
x==Sqrt[x]
x==1 and zero
1==Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]==0
1==0

Your third line incorrectly assumes that any solution to the second line is a solution to the first line. This is not the case.

Consider
x = 7
x * 0 = 7 * 0
0 = 0, which holds for all x (including 2)
2 = 7

[soandos]

I am not multiplying by zero in any step though, i am just saying that Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]==x
means that Sqrt[0+x]==x

[CRGreathouse]

I am not multiplying by zero in any step though, i am just saying that Sqrt[0+Sqrt[0+Sqrt[x+Sqrt[0+...]]]]==x
means that Sqrt[0+x]==x

You are not multiplying by zero, correct. You are saying that line 1 implies line 2, which is a correct statement. You further claim (implicitly) that an x-value solving line 2 solves line 1, which is incorrect.

[soandos]

why does it not follow?

[CRGreathouse]

Why would it?

If someone is 52, then they're in their 50s. If someone is in their 50s, they might be 58. If someone is 58, they are eligible to join the 55+ club. But it doesn't follow that someone who is 52 might be able to join the 55+ club.

If x = -3 then x^2 = 9. If x^2 = 9 it follows that x is 3 or -3. But it does not follow that both 3 = -3 and -3 = -3.

[epkid08]

Just because the zeros of a function equal zero when put through the function, doesn't mean that the zeros equal each other.

(x+1)(x-1)=0
x=1;x=-1
1≠-1

[soandos]

thanks.
btw epkid08, that is not what i was saying. i was saying that value of the function was x, not the zeros of f(x)

[epkid08]

thanks.
btw epkid08, that is not what i was saying. i was saying that value of the function was x, not the zeros of f(x)

It was an analogy. The message was that if x = {a,b}, a≠b.

x=√(x)
x={0,1}
0≠1

[CRGreathouse]

It was an analogy. The message was that if x = {a,b}, a≠b.

x=√(x)
x={0,1}
0≠1

My message was that if F(x) --> G(x), and G(x) holds precisely for x in X, F(x) need not hold for all x in X.