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For each case below; prove that the two given groups are isomorphic(a) R under addition and R under the operation x *y = x+y-4...

Question

For each case below; prove that the two given groups are isomorphic(a) R under addition and R under the operation x *y = x+y-4

For each case below; prove that the two given groups are isomorphic (a) R under addition and R under the operation x *y = x+y-4



Answers

If R is irreflexive and transitive then it's asymmetric. Prove it

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.

And this problem, we're talking about the properties of sets and specifically the relations between sets. We're talking about if a set or relation is reflexive, transitive and symmetric, and of course anti symmetric. The only thing in this problem is that we're not doing this proof based, we have to do this with diagrams. So what does that mean? We're going to have to analyze this problem using matrices. So let's first write our first matrix will call this AM sabar the matrix with respect to our relation. And our matrix would look like this. Who would have these entries? 010111011 So what does this mean? Well, are is going to be reflexive if each element in the diagonal of our matrix is one. And clearly we can see that is not the case. We have this first entry as zero. So our relation is not reflexive. And now let's square our matrix. So we basically take the first matrix and multiply it by that same matrix. When we do that we would get our square matrix equals this 11113 to 1 to two. So what does this mean? Well, the zero entries and our original matrix are non zero in our square matrix. So our isn't not to reflect part of me are is not transitive and I'm moving on to symmetry. How would we do this with matrices? We would have to transpose our original matrix. So are transposed matrix which we denote by this M. R. Raise to this. Tea is the exact same as our original matrix. So we can say that this is definitely symmetric because are transposed matrix is the same as our original now because our relation is symmetric, it's probably not anti symmetric. But we can check it proof based again, this is not anti symmetric because if we have some arbitrary matrix M. Sub I J equals zero, or an arbitrary matrix M sub J I equal to zero. Those don't exist for every I not equal to J. So we're not going to see Auntie Cemetery with with respect to this relation. So I hope this problem helped you understand how we can interpret the properties of the relations on sets, specifically using diagrams and matrices, as opposed to a proof based approach.

Reflexive and transitive than ours? Asymmetric. I think this is the most straightforward of the three Bruce. Okay, so uh let R. B. Your reflexive and transitive. Um oh let X Y bien are and now what I'm going to do is I'm going to suppose that Y X I suppose why X is in our then since our is transitive, X. X is in our and why why is in our but that's a contradiction are is mhm. It reflexive. So Y X. What I supposed right here cannot be in our and thus our is anti symmetric. Oh, no, no, no. Asymmetric, sorry. Yeah.


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