13 on calculator mistakes and estimates

On Calculator Errors and Estimates
All decimal answers in this course are rounded off to three
significant digits. To obtain this,
start from the 1st nonzero digit,
round off to the third digit.
For example, 12.35 ≈12.4
0.001235 ≈ 0.00124.

Calculator Errors

Calculator Errors
As mentioned in the last section that when we calculate with a
machine we can make syntactic and/or semantic mistakes.

Calculator Errors
machine we can make syntactic and/or semantic mistakes. The
syntactic mistakes will be pointed out by the machine.

Calculator Errors
syntactic mistakes will be pointed out by the machine. But the
machine will execute semantic mistakes and return wrong
answers often without our knowledge.

Calculator Errors
answers often without our knowledge. When possible, take the
following steps to reduce semantic mistakes.

Calculator Errors
* Have an estimate of the final outcome. If needed,
use the calculator to come up with an estimate.

Calculator Errors
Example A.
a. Estimate √30, then find its calculator answer.

Calculator Errors
Example A.
Here is a short list of square roots.

A square–root list

Calculator Errors
Example A.
25 < 30 < 36, hence √25 < √30 <√36,


Calculator Errors
Example A.
25 < 30 < 36, hence √25 < √30 <√36, so
5 < √30 < 6.


Calculator Errors
Example A.
25 < 30 < 36, hence √25 < √30 <√36, so
5 < √30 < 6.
Since 30 is about half way between 25 and 36,
so we estimate that √30 ≈ 5.5, half way between 5
and 6. A square–root list

Calculator Errors
Example A.
25 < 30 < 36, hence √25 < √30 <√36, so
5 < √30 < 6.
Since 30 is about half way between 25 and 36,
so we estimate that √30 ≈ 5.5, half way between 5
and 6. In fact √30 ≈ 5.47722… ≈ 5.48 A square–root list

b. Estimate –4 – √27 , then find its calculator answer.
6

6
Using 5 ≈ √27, –4 – √27 ≈ –9,

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
The calculator answer is –4 – √27 ≈ –1.53 which is
6
the approximate, or the numeric answer.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
If a whole Crazy Chicken cost $8.99 then it’s easier to use
$ 9 as an over or upper estimate.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
$ 9 as an over or upper estimate. The real cost of two
chickens must be less than $18 (an upper estimate).

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
The true answer must not be more than any upper estimate.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
If a whole Crazy Chicken and a drink cost $11.08, then $11 is
a lower or under estimate.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
a lower or under estimate. The true cost of two chickens and
two drinks must be more than the lower estimate of $22.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
The true answer must not be less than any lower estimate.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
A semantic error had occurred if the answer falls outside of
the range between a lower estimate an upper estimate.

6
Using 5 ≈ √27, –4 – √27 ≈ –9, so –4 – √27 ≈ 2 .
–3
6
6
A semantic error had occurred if the answer falls outside of
the range between a lower estimate an upper estimate.
Here are some basic about estimating arithmetic calculation.

We note first that the answer for –3+5 can’t be more than 5 nor
be less than –3.

be less than –3.
i. If P is positive and N is negative, then P+N is no more than P,
and no less than N.

be less than –3.
and no less than N.
If there are more terms, the sum of all the positive terms is an
upper estimate and the sum of all the negative terms is a lower
estimate.

be less than –3.
and no less than N.
estimate. So for –23+2(2)2–3,

be less than –3.
and no less than N.
estimate. So for –23+2(2)2–3, +2(2)2 = 8 is an upper estimate

be less than –3.
and no less than N.
and –23–3 = –11 is a lower estimate.

be less than –3.
and no less than N.
Next we note that is 1/3 is smaller than 1/2,

be less than –3.
and no less than N.
Next we note that is 1/3 is smaller than 1/2, i.e. when sharing a
pizza, more people–less pizza (per person), or that
less people–more pizza (per person).

be less than –3.
and no less than N.
ii. Given a fraction N of two positive numbers,
D

be less than –3.
and no less than N.
D
{ if the denominator D increases, the new fraction is less.
if the denominator D decreases, the new fraction is more.

be less than –3.
and no less than N.
D
{ if the denominator D increases, the new fraction is less.
if the denominator D decreases, the new fraction is more.
if the numerator N increases, the new fraction is more.
{ if the numerator N decreases, the new fraction is less.

Example D. a. Given an upper estimate of 4.8 + √34.8
6 + √2.1
Justify it.

6 + √2.1
Justify it.
Any fraction with a larger numerator, and/or a smaller
denominator when compared to 4.8 + √34.8 is more.
6 + √2.1

6 + √2.1
Justify it.
6 + √2.1
Hence by replacing the numerator with a larger one

4.8 + √34.8
6 + √2.1

6 + √2.1
Justify it.
6 + √2.1

4.8 + √34.8 5
6 + √2.1 <

6 + √2.1
Justify it.
6 + √2.1

4.8 + √34.8 5+6
6 + √2.1 <

6 + √2.1
Justify it.
6 + √2.1

4.8 + √34.8 5+6
6 + √2.1 < 6

6 + √2.1
Justify it.
6 + √2.1
Hence by replacing the numerator with a larger one and the
one with a small one, we obtain an upper estimate of 11/6.
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6

6 + √2.1
Justify it.
6 + √2.1
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6
The correct answer must be smaller than this upper estimate.

6 + √2.1
Justify it.
6 + √2.1
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6
If we obtained an answer of 2, we would know 2 is a
wrong answer because 2 is above the upper estimate of 11/6.

6 + √2.1
Justify it.
6 + √2.1
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6
Next we observe 24 < 34 or that for the positive power 4,
the larger base 3 produces larger outcome then with base 2.

6 + √2.1
Justify it.
6 + √2.1
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6
iii. Given bp where p > 0 is a fixed power and b > 0,
if b increases, the outcome increases (is larger),

6 + √2.1
Justify it.
6 + √2.1
4.8 + √34.8 5+6 11
6 + √2.1 < 6 = 6
iii. Given bp where p > 0 is a fixed positive power and b > 0,
if b increases, the outcome increases (is larger),
if b decreases, the outcome decreases (is smaller).

b. Use a calculator to find a lower estimate of (1.87 + 0.08 ) 25.
√7
Justify the estimate.

√7
0.08
The expression is of the form b25 where b = (1.87 + )
√7

√7
0.08
√7
Let’s replace b with a smaller number that is easier to execute.

√7
0.08
√7
By dropping the “extra bits” and keeping just the 1.8 we would
have an smaller and easier base to execute.

√7
0.08
√7
Hence 1.825 ≈ 2,408,865 is a useful lower estimate.

√7
0.08
√7
Now when we execute the original problem and the calculator
returns an answer that’s smaller than 2,408,865, then we know
one or more semantic mistakes had occurred.

√7
0.08
√7
However, because negative exponents means reciprocate,

√7
0.08
√7
we have 2–4 < 3–4 or that for the negative power –4, or that
the larger base 3 produces smaller outcome then with base 2.

√7
0.08
√7
iii. Given bp where p < 0 is a fixed negative power and b > 0,

√7
0.08
√7
iii. Given bp where p < 0 is a fixed negative power and b > 0,
if b increases, the outcome decreases (is smaller),
if b decreases, the outcome increases (is larger),

c. Use a calculator to find a upper estimate of (1.87 + 0.08 ) –5.
√7

√7
0.08
The expression is of the form b–5 where b = (1.87 + )
√7

√7
0.08
√7

√7
0.08
√7
Hence 1.8 –5 ≈ 0.0529 when is a useful lower estimate.

√7
0.08
√7
If a suspicious calculator outcome seems to be too large, try
finding an upper estimates that’s below the outcome to justify
our suspicion that the calculator answer was too large.

√7
0.08
√7
Likewise if the outcome seems to be too small, try find a lower
estimate that’s more than the suspiciously small calculator
answer.

√7
0.08
√7
answer. For complicated calculation, we may estimate parts of
the expression to check and isolate input mistakes.

√7
0.08
√7
answer. For complicated calculation, we may estimate parts of
the expression to check and isolate input mistakes. And finally,
When in doubt, insert ( )’s.

Power Equations and Calculator Inputs
No calculator for part A, B and C.
Exercise A. Estimate the following expressions. Find an upper
estimate and a lower estimate. Justify.
1. √7 2. √10 3. √15 4. √29 5. √47 6. √73
7. √17 + √5 8. √37 +√7 9. √24.5 –√4.2 10. √84.3 –√65.8
Exercise B. Find the positive–term total as an upper estimate
and the negative–term total as a lower estimate. Justify.
11. –2 + 6 – 3 – 11 + 14 12. –2 –(–7) – 2 + 15 + 3
13. 42 – 62 – 23 + 2 14. (–2)(–3) – 4 +10 – 32
15. –22 + 26 – 32 16. –2*33 – 42 + 4*23
17. –4(–2)2 + 6 –(–3)2 18. (–2)*(–3)3 – 42 – 4*(2)3
Exercise C. Estimate the following expressions. Find an upper
estimate and a lower estimate. Justify.
19. 1 + √10 20. –2 – √17 21. –5 + √24
2 3 3
–5.3 – √17.1
22. –√14 + √67 6.8 + √24.8
23. 6 – √4.1 24. 1.1 + √4.1
5

Power Equations and Calculator Inputs
Exercise D. Use a calculator to find lower estimates by using a
smaller positive base.
25. (3.11 + √4.12)5 26. (14.2 – √8.77)3 27. (√50.2 + 2.07)4
8.13 6 29. (√102.8 – 0.3 )7 30. (–√3.9 – √8.9)5
28. (√82.8 + √4.2 )
√9.04
Exercise F. Use a calculator to find upper estimates by using a
larger positive base.
31. (2.91 + √8.92)5 32. (9.89 – √4.11)3 33. (√47.8 + 1.97)6
7.93 7 8.9 8.9
34. (√80.8 + √17 ) 35. (√24.8 + √9.1 )4 36. (√24.8 + √9.1 )4
Exercise G. Use a calculator to find lower estimates by using
larger positive base.
37. (2.91 + √8.92)–5 38. (9.89 – √4.11)–3 39. (√47.8 + 1.97)–6
Exercise H. Use a calculator to find upper estimates by using a
smaller positive base.
40. (3.11 + √4.12)–5 41. (14.2 – √8.77)–3 42. (√50.2 + 2.07)–4

13 on calculator mistakes and estimates

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