By Michael Avidon, math editor
For Students: Performance Expectations (CCSS)
This lesson addresses the following Common Core State Standard (CCSS) for Grade 8:
- 8.NS.A.2: Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
Irrational numbers have decimal expansions that are non-terminating and non-repeating. That π is irrational is a fact that will be assumed without proof. A good approximation could be obtained through a complicated procedure involving the Pythagorean theorem, the distance formula, and the approximation of roots. However, we will simply assume that π = 3.14, to the nearest hundredth.
Numbers in the form √n and ∛ are called radicals. If n is a whole number and not a perfect square, then √n is irrational. If n is a whole number and not a perfect cube, then √n is irrational. All of these numbers therefore have non-terminating, non-repeating decimal expansions. In this lesson, we seek approximations to one decimal place for these numbers, and also estimates for expressions involving these numbers and π.
Approximating radicals with guessing and checking
To approximate √n to the nearest whole number, first find the two perfect squares closest to n. For example, suppose that n = 40. We have
36 < 40 < 49
√36 < √40 < √49
6 < √40 < ___
The number 40 is closer to 36 than to 49, so its square root is closest to among all whole numbers. Compare the whole numbers on a number line.
Note that 40 is about “one third of the way” from 36 to 49. So √40 is about 6 and one third. Thus, it should be about 6.3 or 6.4. Square these to see which is closer to 40:
6.32 = 39.69 and 6.42 = 40.96
These are both close to 40, but 39.69 is closer, so we conclude that √40 = 6.3.
On a number line, it looks like this:
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Example 1
Approximate √75 to the nearest tenth.
Solution:
On a number line, compare 75 to the nearest perfect squares:
It is between 64 = 82 and 81 = 9 2, so √75. The number 75 is about two thirds of the way from 64 to 81, so should be roughly 8 and two thirds. This would make it about 8.6 or _____. Square these numbers and compare the results to 75:
These are both close to 75, but the latter is closer, so √75 ≈ 8.7.
To approximate cube roots, you can follow a similar procedure, but you look for perfect cubes that are closest to the radicand (the number inside the radical sign).
Suppose you wish to approximate ∛45. Compare 45 to the nearest perfect cubes:
It is between 27 = 33 and 64 = ☐☐, so 3 < ∛45 < 4. The number 45 is very close to halfway between 27 and 64, and is slightly closer to 27. So let us try 3.4 and 3.5:
3.43 = 39.304 and 3.53 = 42.875
Both of these are less than 45. We need to try another number:
3.63 = 46.656
This result is the closest to 45, so ∛45 = 3.6.
(The process shown here to approximate the numbers to the nearest tenth uses a roughly proportional relationship: for example, assuming that the halfway point on the input corresponds to the halfway point on the output. But the relationship between a number and its square root, or cube root, is not proportional. So you will sometimes have to try another value to get the best approximation, as in the above example.)
Example 2
Approximate ∛13 to the nearest tenth.
Solution:
Compare 13 to the nearest perfect cubes:
It is between 8 = 23 and _____ = 33, so ____ < ∛13 < ____. The number 13 is about one-fourth of the way from 8 to 27, so let’s try 2.2 and 2.3:
2.23 = 10.648 and 2.33 = 12.167
Both of these are less than 13. We need to try another number:
2.43 = 13.824
This is a tiny bit closer to 13 than 12.167. So ∛13 = 2.4 (but not surprisingly, it is 2.35 to the nearest hundredths place, or midway between 2.3 and 2.4).
Estimating the values of expressions
An approximation of something is an inexact value, but it is very close to the actual value. An estimate is less precise and is best given as an interval. Going back to the first example, 6 < √40 < 7 is an estimate and is an approximation. Technically, an approximation is also an interval. Saying that √40 ≈ 6.3 to the nearest tenth is really saying that 6.25
- <
≈40 < 6.35. But an interval for an approximation is much shorter than an interval for an estimate.
In 6 < ≈40 < 7, the number 6 is called an under-estimate or a lower bound for ≈40. The number 7 is called an over-estimate or an upper bound.
Example 3
The golden ratio is the number ø = 1 + √5 ⁄ 2.
Use √5 a whole-number estimate for to get an estimate for 𝜙. Then approximate the radical to the nearest tenth and use that to get an approximation for 𝜙.
Solution:
Because 4 < 5 < 9, it follows that
____ < √5 < 3. Take square roots
____ < 1 + √5 < 4 Add 1.
3 ⁄ 2 < 1 + √5 ⁄ 2
1.5 < 𝜙 < 2 Simplify
The number 5 is one-fifth of the way from 4 to 9, so approximate √5 with 2.2: 2.22 = 4.84. That is less than 5, so try 2.3: 2.32+5.29. The first square is closer to 5, so √5 = 2.2. This implies 1 + √5 = ______, and therefore 𝜙 = 1.6.
Note that using an approximation to the nearest tenth for part of a more complicated expression, as was done above, does not necessarily give you an approximation to the nearest tenth for the value of the expression. It might be more accurate or less accurate than that. There is a problem in the exercises that addresses this.
Example 4
A circle inscribed in a cardboard square with side length 6 inches is cut out. Using the approximation 3.1 < π < 3.2, find an estimate for the remaining area, with the best possible upper and lower bounds.
Solution:
The area of the square is ____ square inches. The area of the circle is πr2 = π x ____2 = 9π square inches.
The remaining area is thus 36 – 9π square inches. We have
3.1 < π < 3.2
9(3.1) < 9π < 9(3.2)
27.9 < 9π < ______
-28.8 < -9π < -27.9
36 – 28.8 < 36 – 9π < 36 – 27.9
7.2 < remaining area < 8.1
Example 5
Which expression has a greater value: 9√2 or 4π?
Solution:
The number 2 is between 1 and 4 (and closer to 1), so √2 is between 1 and __ (and closer to 1). Now 1.32 = 1.69, 1.42 = 1.96 and 1.52 = 2.25 so √2 is greater than and approximately equal to _____. Therefore, 9√2 is greater than and approximately equal to 12.6. You know that π = 3.14 to the nearest hundredth, so π < 3.145 → 4π < 12.58. Therefore, 4π < 12.58 < 12.6 < 9√2. Thus, 9√2 has the greater value.
Note that in this solution, the following type of comparison was made:
1st expression < upper bound for 1st expression < lower bound for 2nd expression < 2nd expression
If you compare two upper bounds or two lower bounds, you will not be able to combine the inequalities to reach a comparison of the expressions, as was done above (e.g., expression < upper < upper > expression). As a practical matter, you may need to find both bounds for both expressions and then determine which ones to compare in order to reach a conclusion.
Exercises for lesson 2
Determine whole-number estimates and an approximation to the nearest tenth.
- √12
- √18
- √33
- √60
- ∛7
- ∛90
- Which point best represents ? Explain.
Use the estimates and approximations from the previous exercises to determine estimates and approximations for the following expressions.
- 10 – √12
- 2√18 +½
- ∛90 ⁄ 3
Use to get estimates for the following expressions.
- π ⁄ 2
- 1.1π
- volume of a cylinder of radius 1 and height 10
- Which expression has a greater value: 5√5 or 2√33?
- A square with side length 2 is inscribed in a cardboard circle. The square is cut out. Using the approximation , find an estimate for the remaining area, with the best possible upper and lower bounds.
- Challenge Problem:
(a) Use reasoning (and no calculator) to determine how accurate the approximation for 𝜙 in example 3 was.
(b) Write another expression involving √5 such that using the approximation √5 ≈ 22 yields an approximation that is less accurate (not accurate to the nearest tenth).
NEXT LESSON: At-home math lesson: Approximation of pi