Math Homework. Do It Faster, Learn It Better.

Accuracy and Error

Some math problems require an exact answer, while for others, an approximate answer is good enough.

When a problem involves measurement of a real-world quantity, there is always some level of approximation happening. For example, consider two different problems involving measures of time. If the problem involves sprinters running the 100 -meter dash, you will need to use precise measurements: hundredths of a second or better. If the problem involves people’s ages, it’s usually enough to approximate to the nearest year.

Sometimes when a problem involves square roots or other irrational numbers like pi , or even a complicated fraction, it’s useful to use decimal approximations—at least at the end, when you’re reporting your answer.

Example 1:

Deanna is making a wire fence for her garden in the shape below. Find the length of the hypotenuse of the triangle in meters.

Using the Pythagorean Theorem, we get:

x 2 = 3 2 + 7 2 x 2 = 9 + 49 x = 56

So the hypotenuse is 56 meters long. That’s the exact answer. But having the answer in this format isn’t very useful if we’re trying to build something. How do you cut a piece of wire 56 m long?

A good calculator will tell you the value of 56 correct to 30 decimal places:

56 7.483314773547882771167497464633

Even this is an approximation. But it’s a much better approximation that you need. In this case, rounding the value to the nearest centimeter (hundredth of a meter) is probably enough.

56 = 7.48

The accuracy of a measurement or approximation is the degree of closeness to the exact value. The error is the difference between the approximation and the exact value.

When you’re working on multi-step problems, you have to be careful with approximations. Sometimes, an error that is acceptable at one step can get multiplied into a larger error by the end.

Example 2:

A plastic disk is the shape of a circle exactly 11 inches in diameter. Find the combined area of 10 , 000 such disks.

Suppose you use 3.14 as an approximation for π . Using the formula for the area of a circle, we get the area of one disk as:

A = π r 2 = π 11 2 = 121 π 121 ( 3.14 ) = 379.94

Multiply this value by 10 , 000 to get the combined area of 10 , 000 disks.

379.94 × 10 , 000 = 3 , 799 , 400

This gives an answer of 3 , 799 , 400 square inches. But wait!

See what happens when we use a more accurate value for π like 3.1416 :

A = 121 π 121 × 3.1416 = 380.1336

Multiply this by 10 , 000 to get the combined area:

380.1336 × 10 , 000 = 3 , 801 , 336

This is almost 2000 square inches more than our previous estimate!