- #1

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if:

1^x=1^y

and as:

1^2=1^99

then:

1=99

1^x=1^y

and as:

1^2=1^99

then:

1=99

- Thread starter nabodit
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- #1

- 16

- 0

if:

1^x=1^y

and as:

1^2=1^99

then:

1=99

1^x=1^y

and as:

1^2=1^99

then:

1=99

- #2

arildno

Science Advisor

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Why do think this is true?

Since 1*0=99*0, do you consider this as proof of your assertion as well?

Since 1*0=99*0, do you consider this as proof of your assertion as well?

Last edited:

- #3

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The log of your equation would beg to differ... EDIT: except you consistently used 1, not a variable. Duh! Mondays are rubbish. Anyway, shouldn't that read 2 = 99?

Last edited:

- #4

VietDao29

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I dunno if it's a joke or not.

If you have a function f(x) such that [itex]f(x_0) \neq f(x_1)[/itex], for all [itex]x_0 \neq x_1[/itex]. Just in that case, you will have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]

Some function like :f(x) = 0x, f(x) = 1 ^ x, f(x) = x ^ 0 ([itex]x \in \mathbb{R} - \{ 0 \}[/itex]). You cannot have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]. Why? Because in that 3 examples:

[itex]\forall x, f(x) = const[/itex]

Viet Dao,

If you have a function f(x) such that [itex]f(x_0) \neq f(x_1)[/itex], for all [itex]x_0 \neq x_1[/itex]. Just in that case, you will have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]

Some function like :f(x) = 0x, f(x) = 1 ^ x, f(x) = x ^ 0 ([itex]x \in \mathbb{R} - \{ 0 \}[/itex]). You cannot have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]. Why? Because in that 3 examples:

[itex]\forall x, f(x) = const[/itex]

Viet Dao,

Last edited:

- #5

Alkatran

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Oh no! 2 pi = 0, meaning pi = 0, which means circles don't exist!

If only there was a flaw in the logic...

- #6

Gokul43201

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LHS = 1*1 = 1nabodit said:1^2=1^99

then:

1=99

RHS = 1*1*1* ...(ninety five times) *1 = 1

LHS = 1 = RHS

How do you go from line (1) to line (2) ?

nabodit : If you have a question to ask, ask it now.

- #7

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Reading this really brightened up my day. :rofl: