Is this logical reasoning correct?

  • Thread starter frb
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  • #1
frb
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Suppose I want to show that two functions f and g are equal. A way to prove this could be to prove the statement:
[tex] f(x) = n \Leftrightarrow g(x) = n[/tex]

Is it enough to show one side of the implication?
Prove the following statement:
[tex] f(x) = n \Rightarrow g(x) = n[/tex]

and reason as follows, suppose [tex] f(x) \neq n[/tex],
[tex] \Rightarrow \exists m\neq n: f(x)=m[/tex]
[tex]\Rightarrow g(x)=m\neq n[/tex]
Which would mean that I have shown the converse implication, and thus I have equivalence.
 
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  • #2
tiny-tim
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Prove the following statement:
[tex] f(x) = n \Rightarrow g(x) = n[/tex]

and reason as follows, suppose [tex] f(x) \neq n[/tex],
[tex] \Rightarrow \exists m\neq n: f(x)=m[/tex]
[tex]\Rightarrow g(x)=m\neq n[/tex]
Which would mean that I have shown the converse implication, and thus I have equivalence.
Hi frb! :smile:

Sorry, but I don't understand your proof, so I'm going to say no, it's not correct. :redface:
 
  • #3
frb
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Suppose I have proven the statement f(x) = n implies g(x)=n.

To obtain equivalence I have to prove f(x) != n implies g(x) != n.

So i reason as follows, suppose f(x) != n,
f(x) surely has another value, let this value be m, so f(x) = m, and m != n.
f(x) = m implies g(x) = m, and that implies that g(x) != n.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Suppose I have proven the statement f(x) = n implies g(x)=n.

To obtain equivalence I have to prove f(x) != n implies g(x) != n.

So i reason as follows, suppose f(x) != n,
f(x) surely has another value, let this value be m, so f(x) = m, and m != n.
f(x) = m implies g(x) = m, and that implies that g(x) != n.
ah … that seems fine …

but if you want to prove f = g, which is the same as, for all x, f(x) = g(x),

then if you know for all x, f(x) = n implies g(x)=n,

then for all x, f(x) = g(x) … you don't need any more logic than that. :smile:
 
  • #5
frb
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of course. My brain is fried due to too much studying. I tend to make things more difficult then. I thought that there should be a flaw or something. Thanks though!
 
  • #6
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This proof assumes that f and g have the same domain.

It fails if they have different domains. But then by definition the functions are not equal, so there was no need to go any further.
 

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