Show its not a group for # where a#b=a+b-ab in the set of all real numbers

In summary, the task is to show that the system S = {set of all real numbers ( \Re )}, where a#b=a+b-ab, is not a group. This is followed by finding a value of c such that the system = {\Re \cap \overline{c}, # } is a group, using the definition of a group which includes being closed, associative, having an identity, and each element having an inverse. After attempting to show that the system is closed, has an identity, and has inverses, there is confusion about its associativity.
  • #1
brandy
161
0

Homework Statement


In set theory, i have a two part question, the first is showing that the system S={set of all real numbers( [itex]\Re[/itex] )}, #}
where a#b=a+b-ab
we have to show that it's not a group.
and then find what c is so that the system = { [itex]\Re[/itex] [itex]\cap[/itex][itex]\overline{c}[/itex], # } is a group.

Homework Equations


the definition i have of a group is closed, associative, has an identity and each element has an inverse.

The Attempt at a Solution


i figured it has to be closed, as in a,b,a#b [itex]\in[/itex] set of all real numbers
identity: i did a#e-ae=0 ->e(1-a)=0,e=0
inverse: a#b=e=a+b-ab, a+b-ab=0 ->a=-b/(1-b), then you have b-b/(1-b)+b^2/(1-b) which is equal to 0 so it works.
so far so good.
associative: i checked it and unless i made a mistake (a#b)#c=a#(b#c), it looks commutive so i tried (a#b)#(c#d) verses a#(b#c)#d but unless i made a mistake this wasn't equal. The system didnt look like it should be associative but by using the definition we were given, it was.
so I'm confused. can someone help point me in the right direction?
 
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  • #2
a=-b/(1-b),

I can think of a value of b for which this doesn't work
 
  • #3
*facepalm*
thanks!
 

FAQ: Show its not a group for # where a#b=a+b-ab in the set of all real numbers

1. What is the purpose of this group?

The purpose of this group is to study the operation #, defined as a#b=a+b-ab, in the set of all real numbers.

2. How is the operation # different from addition and multiplication?

The operation # is different from addition and multiplication because it combines both operations in a unique way, resulting in a different output.

3. Can you give an example of how # works?

Sure, for example, let's take a=2 and b=3. Then, 2#3=2+3-(2*3)=2+3-6=-1. So, 2#3=-1 in the set of all real numbers.

4. Is the operation # commutative?

No, the operation # is not commutative. This means that the order in which the numbers are used in the operation affects the result. For example, a#b does not necessarily equal b#a.

5. How is this group relevant to real-world applications?

This group may not have direct real-world applications, but it helps in understanding how different operations can be combined and how the order of operations can affect the result. This can be helpful in solving more complex mathematical problems and in computer programming.

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