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## Ask Professor Puzzler

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Sixth grader Larry asks, "Is there distributive property in Star Operations...."

Good morning, Larry. Before answering your question, I'd like to give a bit of background on Star Operations, in case someone is not familiar with them. If you're not interested in background, you can skip to the horizontal bar, where I'll try to answer your question.

Star Operations are arbitrarily constructed operations in which a definition is given for how the operation is performed between two variables. They are referred to as "star operations" because the asterisk is often used as the symbol for the operation. For example, here is a star operation definition:

a * b = a2 + b2.

I had a student ask me once what these operations are good for, and my answer was, "They're good for teachers who like to torment students." That was, of course, a bit tongue-in-cheek, but the primary purposes they serve are:

• Giving students practice with arithmetic/algebraic manipulation
• Introducing the concept of functions, without actually using the word or symbolism of functions. The star operation given above can also be written as:

f(a, b) = a2 + b2

With that as background, let's get to Larry's question.

For the sake of this discussion, let's use the star operation given above: a * b = a2 + b2. Larry wants to know if the distributive property works for star operations, so let's give it a try and see what happens. We kind of need to guess how a distributive property would be implemented - there are several possibilities, which I'll list below:

1. a(x * y) = ax * ay
2. a * (x + y) = a * x + a * y
3. a * (x * y) = (a * x) * (a * y)

In the first case, the addition was replaced with a star. In the second, the multiplication was replaced with a star, and in the third, both the addition and the multiplication were replaced with a star. Let's test each of these by plugging in values for a, x, and y. If the values we pick work, that doesn't prove that the rule is legitimate, but if the values don't work, we know that the rule is bogus. Let's do the following values:

a = 3; x = 4; y = 2.

## Possibility #1

3(4 * 2) = 3(4) * 3(2)
3(16 + 4) = 12 * 6
3(20) = 144 + 36
60 = 180, which is NOT true, so the first possibility is bogus

## Possibility #2

3 * (4 + 2) = 3 * 4 + 3 * 2
3 * 6 = (9 + 16) + (9 + 4)
9 + 36 = 25 + 13
45 = 38;  also bogus!

## Possibility #3

3 * (4 * 2) = (3 * 4) * (3 * 2)
3 * (16 + 4) = (9 + 16) * (9 + 4)
3 * 20 = 25 * 13
9 + 400 = 625 + 169
409 = 794; BOGUS!

So we tried all the possibilities, and none of them worked. That doesn't mean that they are bogus for all star operations. However, it doesn't work for all star operations, it's not really good for much, is it?

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