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corey1234
 3 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(xy). Now when I do it, I break up the absolute value into xe for xe greater than 0 and (xe) for xe less than 0, where e is the identity element. Please explain your steps to this problem.
corey1234
 3 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(xy). Now when I do it, I break up the absolute value into xe for xe greater than 0 and (xe) for xe less than 0, where e is the identity element. Please explain your steps to this problem.

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jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1x * y = x  y x * e = x  e x = x  e  If x  e > 0, then x = x  e xx = e 0x = e 0 = e e = 0 e = 0  If x  e < 0, then x = (x  e) x = x  e xx = e 2x = e e = 2x e = 2x So the identity element is either e = 0 or e = 2x for any real x.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1I guess I should say though that if you're going for one unique identity element, then it would simply be 0

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.12x 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1i 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes, but we found that e = 2x and I showed that x = 0 so naturally e = 2x > e = 2*0 > e = 0

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1x * 0 = x for all x (when x > 0 or x < 0)

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0Ok 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==xy. We check xe, which equals xe when x>e and (xe) when x<e. If x>e, then xe==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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1when 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0Ok so what you are saying is that e*x== ex when x<e and xe when x>e and since e==0 and x>0 , then 0*x== x0 which is just x?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes, sry i was trying a different way to do this, but not really finding anything

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0Also, when we first said that x>0, does that mean that that condition has to be true in order for x0=x? Because x can be negative there?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0yes but then x*e==(xe)=x works when x is negative if e==0. For example x0=x, x can be negative there.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1you 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0But why do you have to go back to the definition everytime. I mean, you used the fact that xe==(xe)=x to solve for e. And you found out that e=0. So shouldn't that condition hold for x0=x? But it doesn't since x can be negative there.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1you want to find an identity element (e) for x * y = x  y not for x * y = x  y

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1we break down xy 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1then we check to see if e = 0 is actually the true identity element for x * y = x  y

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0So what you are saying is that x>0 is only a condition for xe to equal (xe), and the fact that xe=x is something totally different to just find e, which equals 0?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes x > 0 means x  0 = x  0, but we want something that will apply to both x > 0 and x < 0

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0But 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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1good 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0So then there is no identity element for the set of real numbers for this operation.?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes it looks like it since it doesn't apply to all real numbers

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1personally, I would restrict the domain and say e = 0 is the identity element, but it looks like there isn't one here

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0It just doesn't make sense to me how x>0 for xe=xe but x can be nagative for xe=x. Then there is a contradiction right?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yeah 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1one 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

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0But what about xe==x. That can have negative x values. And x>0. Is that what you are saying?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes x  e = x can have negative values, but we're referring to xy not xy

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1this is why it's very important to check the possible solutions

corey1234
 3 years ago
Best ResponseYou've already chosen the best response.0Ok thank you so much.
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