## corey1234 2 years ago Ok. I have a very hard question. Hopefully someone here can solve it. Now for binary operations, well I have one here, and I need to know if there exists in R (the set of real numbers), an identity element for the operation *. Here is the operation, x*y=abs(x-y). Now when I do it, I break up the absolute value into x-e for x-e greater than 0 and -(x-e) for x-e less than 0, where e is the identity element. Please explain your steps to this problem.

1. jim_thompson5910

x * y = |x - y| x * e = |x - e| x = |x - e| ------------------------------------------------------- If x - e > 0, then x = x - e x-x = -e 0x = -e 0 = -e -e = 0 e = 0 ------------------------------------------------------- If x - e < 0, then x = -(x - e) -x = x - e -x-x = -e -2x = -e -e = -2x e = 2x So the identity element is either e = 0 or e = 2x for any real x.

2. jim_thompson5910

I guess I should say though that if you're going for one unique identity element, then it would simply be 0

3. corey1234

Yes, but the definition of an identity element for R is that x*e==e*x==x for ALL of R. The e=0 is only true for x greater than 0 and the e=2x is only true for x greater than 0 too. That is the confusing part.

4. jim_thompson5910

2x is only an identity for x, so 2x doesn't work for all x 0 works for all x so 0 is an identity of x*y

5. jim_thompson5910

i should clarify, e = 2x only satisfies x * e = 0 for any x since it doesn't work for e * x = 0, it starts to fall apart

6. jim_thompson5910

if you repeat the steps above, but do it for e * x instead of x * e, you'll find that e = 2x when e - x > 0 so e = 0 when x - e > 0 and e = 2x when e - x > 0 this means 2x = 0 ----> x = 0

7. corey1234

I am not so sure what you are talking about. You are not trying to solve for x. You are trying to solve for e. Now for x>e then e==0 and for x<e, e==2x. But when e==2x, then that means x<2x which means x>0. aND X*0==X AND X* 2X ==-X. So e=2x is automatically out of the question. But for e=0, x*0 only equals x when x>0.

8. jim_thompson5910

yes, but we found that e = 2x and I showed that x = 0 so naturally e = 2x ---> e = 2*0 ----> e = 0

9. jim_thompson5910

x * 0 = x for all x (when x > 0 or x < 0)

10. corey1234

Ok I am kind of confused now. Can we start this from the beginning. So we are looking to see if there is an identity element for x*y==|x-y|. We check |x-e|, which equals x-e when x>e and -(x-e) when x<e. If x>e, then x-e==x and e==0. So that means if x>0, then x*0==x and 0*x==-x. This is a problem. e=0 only works as an identity element when x*0 not 0*x. Now we check for x< e. Then we agree that e==2x. So x*2x== -x for x>0 and x for x<0. So again, it is only x for when x is less than 0. So this is not an identity element either.

11. jim_thompson5910

when you say "e=0 only works as an identity element when x*0 not 0*x. " that's just not true IF x * y = x - y, then it would be true BUT the binary operation is x * y = |x - y| so x * 0 = |x - 0| x * 0 = |x| x * 0 = x and 0 * x = |0 - x| 0 * x = |-x| 0 * x = x So 0 is clearly an identity element for x * y = |x - y|

12. corey1234

Ok so what you are saying is that e*x== e-x when x<e and x-e when x>e and since e==0 and x>0 , then 0*x== x-0 which is just x?

13. jim_thompson5910

yes, sry i was trying a different way to do this, but not really finding anything

14. corey1234

Also, when we first said that x>0, does that mean that that condition has to be true in order for x-0=x? Because x can be negative there?

15. jim_thompson5910

yes you have to force x to be positive because this example below shows where things break down 0 * x = |0 - x| say x = -2 0 * (-2) = |0 - (-2)| 0 * (-2) = |0 + 2| -2 = 2

16. corey1234

yes but then x*e==(x-e)=x works when x is negative if e==0. For example x-0=x, x can be negative there.

17. jim_thompson5910

you have to go back to the definition of what the binary operator is everytime it works for x - 0 = x but it doesn't work for |x - 0| = x

18. corey1234

But why do you have to go back to the definition everytime. I mean, you used the fact that |x-e|==(x-e)=x to solve for e. And you found out that e=0. So shouldn't that condition hold for x-0=x? But it doesn't since x can be negative there.

19. jim_thompson5910

you want to find an identity element (e) for x * y = |x - y| not for x * y = x - y

20. jim_thompson5910

we break down |x-y| into its smaller parts to find the various possible solutions, then we piece them back together to find that only e = 0 is the identity element

21. jim_thompson5910

then we check to see if e = 0 is actually the true identity element for x * y = |x - y|

22. corey1234

So what you are saying is that x>0 is only a condition for |x-e| to equal (x-e), and the fact that x-e=x is something totally different to just find e, which equals 0?

23. jim_thompson5910

yes x > 0 means |x - 0| = x - 0, but we want something that will apply to both x > 0 and x < 0

24. corey1234

But all of this works only when x>0. Which means e=0 is not an identity element because it is not true for all x in R.

25. jim_thompson5910

good point, if you restrict the domain to the set of positive numbers, then it works if the domain is the set of all real numbers, then it won't work because of the negative numbers

26. corey1234

So then there is no identity element for the set of real numbers for this operation.?

27. jim_thompson5910

yes it looks like it since it doesn't apply to all real numbers

28. jim_thompson5910

personally, I would restrict the domain and say e = 0 is the identity element, but it looks like there isn't one here

29. corey1234

It just doesn't make sense to me how x>0 for |x-e|=x-e but x can be nagative for x-e=x. Then there is a contradiction right?

30. jim_thompson5910

yeah in one sense, we're saying e = 0 is an identity element (which is supposed to be unique), but in another, we're saying e = 2x which says that we have infinitely many identity elements

31. jim_thompson5910

one quick way to determine there is no identity is this x * e = |x - e| x = |x - e| which only has solutions when x > 0 like you said (since k = |x| only has solutions when k > 0) but x is allowed to be negative, which contradicts this fact so there are no identity elements

32. corey1234

But what about x-e==x. That can have negative x values. And x>0. Is that what you are saying?

33. jim_thompson5910

yes x - e = x can have negative values, but we're referring to |x-y| not x-y

34. jim_thompson5910

this is why it's very important to check the possible solutions

35. corey1234

Ok thank you so much.

36. jim_thompson5910

np