Is Jon's Definition of Invertibility Correct for Functions Between Sets A and B?

  • Thread starter Thread starter Seda
  • Start date Start date
  • Tags Tags
    Relations
Seda
Messages
70
Reaction score
0

Homework Statement



A.) Jon wants to define a function f: A->B as invertible iff for all a in A and all b in B with f(a)=b, there exists a function g:B->A for which g(b)=a.

Is that reasonable?


B.) Determine Whether the relation ~ on the Real Numbers defined by x~y is reflexive, symmetric, or transitive.

1.) x~y iff xy<= 0
2.) x~y iff xy < 0


Homework Equations



None really, except maybe a definition for invertible.

The Attempt at a Solution


this seems to make sense, but it seems odd to answer a math question with a "yes" and move on. Am I missing something about the defininition of invertibility that makes the statement in the question incorrect?

For B, these questions seem really easy, but they also seem to be exactly the same. Both relations seem to be Symmetric only...because x^2 is not less than zero for all real values, and the counterexample x=-1, y=1, z=-1 proves that both aren't transitive. AM i missing something?
 
Physics news on Phys.org
A) looks a little subtle. Think about it. For one thing the quantifiers smell wrong. You said for all a and b there exists a function g. Jon didn't say that the g should be the same for ALL choices of a and b. Second, worry about the case where f isn't onto (surjective). What is your definition of 'invertible'? B) looks pretty reasonable to me.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top