The asterisk (*) and pound (#) symbols

[KCA_Math]

I had math through Diff Eq at Ga Tech, but that was 27 years ago. I'm now helping a high school student and have been able to work every problem he didn't understand save one. I honestly don't remember this type problem or symbology at all. First, I believe the asterisk is not a multiplication symbol in this problem. It follows:

If a * b = a - 3b and a # b = 2a + 3b, evaluate (4 * 3) # 5

The solution says to replace (a * b) with (4 * 3) to give (4 * 3) = a - 3b. a is assumed to be 4 and b is assumed to be 3. [Why?]
So we have (4 * 3) = 4 - 3(3) = -5. Because we know (4 * 3) = -5 we can now evaluate -5 # 5.
We can now replace a # b with -5 # 5 to get -5 # 5 = 2a + 3b = 2(-5) + 3(5) = 5. [Now a,b are -5,5 and not 4,3. What?]
So (4 * 4) # 5 = 5

I have no recollection of this and frankly, I don't get the purpose of the exercise. What am I missing? Can someone explain what this is called and possibly point me to a web site where I can understand this in more detail?

Thanks!

 

[KCA_Math]

After an eternity Googling and surfing other math sites I eventually found a discussion of the exact problem here: http://schools.mylounge.com/archive/ind ... 80351.html

I understand the solution, but still don't get the purpose of the exercise. I don't see any related problems anywhere that explain the utility of the problem. I thought it was some kind of "new math" that I missed.

Nonetheless, I'm glad I found this site. It looks like a friendly group of people who love math. I'll fit right in!

 

[Deleted member 4993]

After an eternity Googling and surfing other math sites I eventually found a discussion of the exact problem here: http://schools.mylounge.com/archive/ind ... 80351.html

I understand the solution, but still don't get the purpose of the exercise. I don't see any related problems anywhere that explain the utility of the problem. I thought it was some kind of "new math" that I missed.

Nonetheless, I'm glad I found this site. It looks like a friendly group of people who love math. I'll fit right in!

This is prelude to "abstract algebra" - where new relationships are defined.

It is similar to when in decimal system we define

ab = a*10 + b

 

[mmm4444bot]

I see this as an exerise in symbolic-reasoning skills and logic.

I've seen similar exericses in both intermediate algebra and precalculus courses, albeit rarely.

 

[daon]

Alternatively, you could simplify the complicated "formula" say with a,b,c.

(a*b)#c = (a-3b)#c = 2(a-3b)+3c = 2a-6b+3c

Then plug in a=4,b=3,c=5.

 

[KCA_Math]

Thanks for the replies. I see the purpose of the exercise, but don't remember this type of problem. Of course, there are a lot of things I don't remember! Thanks again.