is this a sufficient proof....

[Semidevil]

this is my attempt at solving this proof:

let function f: R-->R be a real valued function. f is said to be strictly increasing if whenever x < y, then f(x) < f(y).

Prove: If f is striclty increasing, then f is injective.

so here is my take at this. I decide to use proof by contrapositive: If f is not injective, then it is not strictly increasing.

so "not striclty increasing" means that x >= y, which means that f(x) >= f(y).

"not injective" means that f(x) != f(y), which implies x != y....

so this means that If f is not injective, then it is not strictly increasing.

so that is as far as I got w/ my proof............is that sufficient?

if not, where did I go wrong?

 

[FoBoT]

😕

 

[Semidevil]

....uh...bump??

math guru's please help...this is not a hw....just part of study guide for final...

 

[ndee]

you have to multiply f(y) with x^y to get the nut of the tree.

 

[dighn]

Originally posted by: Semidevil
this is my attempt at solving this proof:

let function f: R-->R be a real valued function. f is said to be strictly increasing if whenever x < y, then f(x) < f(y).

Prove: If f is striclty increasing, then f is injective.

so here is my take at this. I decide to use proof by contrapositive: If f is not injective, then it is not strictly increasing.

so "not striclty increasing" means that x >= y, which means that f(x) >= f(y).

"not injective" means that f(x) != f(y), which implies x != y....

so this means that If f is not injective, then it is not strictly increasing.

so that is as far as I got w/ my proof............is that sufficient?

if not, where did I go wrong?

not strictly increasing means there exists x and y such that x < y and f(x) >= f(y) by proper negation of the universal qualifier. what you have doesn' t make much sense.

not inject give means there exists x and y such that x != y and f(x) = f(y) by negation, so that means x < y and f(x) = f(y) or y<x and f(y) = f(x) and either situation violates the definition of strictly increasing

 

[eakers]

assume that f is not injective.
then there exists an x and y where f(x)=f(y) and x =! y
but by hypothesis f is strictly increasing so this is impossible

therefore you have a contradiction so f must be injective.

 

[ndee]

sheesh ladies, stop talkin' so smart.

 

[guyver01]

Originally posted by: eakers
assume that f is not injective.
then there exists an x and y where f(x)=f(y) and x =! y
but by hypothesis f is strictly increasing so this is impossible

therefore you have a contradiction so f must be injective.

:Q

 

[eakers]

Originally posted by: guyver01

Originally posted by: eakers
assume that f is not injective.
then there exists an x and y where f(x)=f(y) and x =! y
but by hypothesis f is strictly increasing so this is impossible

therefore you have a contradiction so f must be injective.

:Q

i'm a math student. this is what i do. 😛