**A BRICK and A HALF**

Jeffrey T. Kertscher

Q: If a brick weighs 3 pounds plus half a brick, how much does a brick and a half weigh?

This is a very interesting problem, because the wording of the question could very easily be interrupted in several different ways.

My first instinct was to yell, "Oh that is easy, the weight of one brick is 3 1/2 lb., so a brick and a half will weigh 5 1/4 lb....Right?" The first problem I had was that I did not read the problem fully, slowly, and carefully. I just skimmed over the high points of the problem, and then found out what the problem asked for. To me the problem read that a brick weighs 3 lb. plus 1/2 of a pound, thus a brick will weigh 3 1/2 lb. I understand that all of the mathematicians are laughing at me now, but all of the math educators realize that I have made a common mistake that is made every day by students in any level of education.

Well, I have spent 5 years at a major university and if there is one thing that I have learned, it is to always go back and check your work (also since I was going to write an essay on this topic, I thought that it was best to make sure that I had the correct answer). After re-skimming the problem I came up with an astonishing revelation...I was wrong. "Well the problem says that a brick weighs 3 lb. and half of a brick is 1 1/2 lb. so a brick and a half must weigh 4 1/2 lb....Right?" From this point I proceeded to make a beautiful spread sheet that used colored cells as manipulative to illustrate this solution. Then ideas of how I could expand the problem into addition, subtraction, multiplication, and division of fractions could be illustrated with similar manipulative, ran through my head.

As I began to write this essay I thought that it would be best if I re-read the problem one more time. Much to my amazement, I had misinterpreted the problem again. The problem is saying that a brick weighs,3 lb.+1/2 of the total brick weight. After reading the problem three times I just now realized the parameters of the problem, and this can be attributed to the fact that I skimmed over the problem carelessly. I have effectively accomplished 2 things, one I have amused several mathematicians with my ignorance, and I realized that when beginning this problem I was in the frame of mind of a typical student.

Let's first discuss the solution to the problem, and then explore more into how a student would view this problem. Once I finally realized what the problem was asking for I started to solve the problem algebraically. In the following equation I will let x = the weight of the brick:

x = 3 lb. + 1/2(x) multiply the equation by 2

2x = 6 lb. + x subtract x from both sides

x = 6 lb. So the weight of one brick is 6 lb. and the weight of a brick and a half of a brick is 9 lb. Our finding can be confirmed graphically using Algebra Xpresser as shown below.

The two lines that were graphed were y = x and y = 3 + 1/2(x), and as you can see they intersect at the point were x = 6. This confirms that the answer to the question of how much one brick weights.

As I began to explore the question I wondered how much the brick would weight if the problem read: A brick weighs 4 lb. plus half a brick..... and in a similar fashion a algebraic and graphic calculations, I found that one brick would weigh 8 lb. This is the beginning of a trend that showed that the weight of a brick was twice as much as the given value. The hypothesis held true for several other random values that I substituted into the equation. So, as any good Math Education student would do I began working on a deductive proof.

Proof:

Assume a is an element of the reals, and x is the weight of the brick. I will prove that if

x = a + 1/2(x) then

x = 2a. So let:

x = a + 1/2(x) Multiply both sides by 2 (multiplication is well-defined in the reals)

2x = 2a + x -add a -x to both sides (additive inverse, well-defineness of addition)

x= 2a Therefore, the proposition holds true. QED

Well I hope at this point I have made all of a the mathematicians happy with my proof, but unfortunately a typical high school student may not comprehend a proof at first. So, what are some other ways to look at and explain this problem. First of all the problem's wording is misleading as mentioned before. Another way to rephrase it would be to say that the weight of one brick is equal to the weight of half a brick plus three pounds. This example takes the focus off the weight of the brick being 3 lb. and puts is more on the weight of half of a brick. This can be written as a mathematical equation:

weight of one brick = weight of half a brick + 3 lb. For this problem to be explained a teacher might approach it in the following way. "Let's assume that we have a brick with a certain weight. Now we are going to cut the brick in half, and this will give us two halves that weigh an equal amount. The sum of the two halves will give us the weight of the total brick. It is given in the problem that the weight of one part is 3 lb., thus the weight of the other part must equal 3 lb. also. So we can see that the weight of one brick would be 3 lb. plus 3 lb. giving us a 6 lb. brick. In order to find the weight of a brick and a half of a brick we must add 6 lb. and 3 lb. giving us 9 lb. as a solution."

Looking at the problem in this ways makes the mathematics simple. The only concept that a student would need to grasp is that if a brick is divided in half, both of the halves will have the same weight. Manipulatives would be very helpful to illustrate this point. For example, if a teacher held up a half of a brick and said that it weighted 3 lb. and then held up another half of a brick and said that it weighed half of a brick. From here the students could see that the two halves would make a whole, and that the halves are equal. A spread sheet can be used to illustrate this point, where the cells are used as manipulative. For example:

This is an outstanding problem to use in a classroom, because it will help to teach the students to read the problem carefully. The first obstacle of the problem is that it is difficult to understand the parameters of the problem. Many students will skim over the problem similar to what I did, come up with a quick solution and then be done with it. After the actual problem is understood the mathematics behind it is not too involved as long as the students are familiar with using variables to represent certain unknowns. Also, the students must be comfortable enough with the algebra to set up the problem. This is an excellent exercise to use in the classroom, not just for the mathematics, but for the problem solving.