- #1

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## Main Question or Discussion Point

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

x==Sqrt[x]

x==1 and zero

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

1==0

- Thread starter soandos
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- #1

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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

x==Sqrt[x]

x==1 and zero

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

1==0

- #2

CRGreathouse

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

Consider

x = 7

x * 0 = 7 * 0

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

2 = 7

- #3

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means that Sqrt[0+x]==x

- #4

CRGreathouse

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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.

means that Sqrt[0+x]==x

- #5

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why does it not follow?

- #6

CRGreathouse

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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.

- #7

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(x+1)(x-1)=0

x=1;x=-1

1≠-1

- #8

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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)

- #9

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

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)

x=√(x)

x={0,1}

0≠1

- #10

CRGreathouse

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

x=√(x)

x={0,1}

0≠1

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