Understanding Inverses & Max* for Integers

  • Thread starter Thread starter kathrynag
  • Start date Start date
  • Tags Tags
    Integers Max
kathrynag
Messages
595
Reaction score
0

Homework Statement


If I have a group defined on the integers, by a*b=ab, how do I know if an inverse exists?
Also, define * on the integers by a*b=max{a,b}




Homework Equations





The Attempt at a Solution


I got 1/a as an inverse, but I'm thinking it's not a group since we don't know if 1/a is an element of the integers.

The max is confusing me.
I know I need to check associativity, identity, and inverse for a group.
a*(b*c)=a*max{b,c)
=max{a,max{b,c}}
(a*b)*c=max{a,b}*c
=max{max{a,b},c}
I'm already confused at this point.
 
Physics news on Phys.org
you only need contradict one of the properties for it not to be a group

The first one is correct, for a not equal to e, the inverse does not exist within the group.

For the 2nd
associativity is ok, as max{max{a,b},c} = max{a,b,c}
you could try the identity, whici si clearly not unique
or probably more straight forward show there is no inverse, consider a*b with a<b
 
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