noko

[Anonymous No.2928759]

dear /sci/ I am trying to do math homework and am stuck please help.

Let f:(all reals)→(all reals). Then f is strictly increasing provided f(a)<f(b) wherever a<b. Prove that if f is strictly increasing, then f is one to one and f inverse is strictly increasing.

Please and thank you.

[Anonymous No.2928763]

[Anonymous No.2928766]

This only works if f is continuous.

[Anonymous No.2928787]

[Scientist !!ThFjnJh4EkH No.2928788]

>>2928766
Don't think so.

Also, OP, it's pretty straightforward proof(s), but it sounds like homework, so I'll pass.

[Anonymous No.2928791]

[Anonymous No.2928792]

[Anonymous No.2928797]

[Anonymous No.2928800]

[Anonymous No.2928802]

[Anonymous No.2928805]

>>2928788
f(x)=x if x<0
f(x)=x+1 if x≥0
is strictly increasing, but it doesn't have a strictly increasing inverse.

[Anonymous No.2928808]

>>2928759
Standard contra-positive proof. Start with your definitions and work it through to you get the result.
>>2928766
We already know it is by...
><span class="math">\Re \Rightarrow \Re [\math][/spoiler]

[Scientist !!ThFjnJh4EkH No.2928815]

>>2928805
What?

I guess it depends on the definition of strictly increasing.

For a function f : A -> B, A subset R, B subset R, f is strictly increasing iff
(for all x in A)(for all y in A)(x < y implies f(x) < f(y))

If that's the definition, then f inverse is strictly increasing. Of course, domain of f inverse is not R.

[Anonymous No.2928817]

>>2928800
it kinda works for bill

[Anonymous No.2928818]

Invert the function and apply a generic one-to-one proof across the board

Protip: convert f to its logical form and apply standard inverse principles

Obviously homework so I'm going to pass on giving any more tips. See enough people in my extra help tutorials that cheesed their way through their second year mathematical proofs courses to willingly give out answers.