math41QA

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course Mth 151

4/5/2012 2:11PM

018. Base-10 Place-value Number System

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Question: `q001. There are 5 questions in this set.

From lectures and textbook you will learn about some of the counting systems used by past cultures. Various systems enabled people to count objects and to do basic arithmetic, but the base-10 place value system almost universally used today has significant advantages over all these systems.

The key to the base-10 place value system is that each digit in a number tells us how many times a corresponding power of 10 is to be counted.

For example the number 347 tells us that we have seven 1's, 4 ten's and 3 one-hundred's, so 347 means 3 * 100 + 4 * 10 + 7 * 1.

Since 10^2 = 100, 10^1 = 10 and 10^0 = 1, this is also written as

3 * 10^2 + 4 * 10^1 + 7 * 10^0.

How would we write 836 in terms of powers of 10?

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Your solution:

8 * 10^2 + 3 * 10^1 + 6 * 10^0

confidence rating #$&*: 3

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Given Solution:

836 means 8 * 100 + 3 * 10 + 6 * 1, or 8 * 10^2 + 3 * 10^1 + 6 * 10^0.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q002. How would we write 34,907 in terms of powers of 10?

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Your solution:

3 * 10,000 + 4 * 1,000 + 9 * 100 + 0 * 10 + 7 * 1 or 3 * 10^4 + 4 * 10^3 + 9 * 10^2 + 0 * 10^1 + 7 * 1^0

confidence rating #$&*: 3

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Given Solution:

34,907 means 3 * 10,000 + 4 * 1000 + 9 * 100 + 0 * 10 + 7 * 1, or 3 * 10^4 + 4 * 10^3 + 9 * 10^2 + 0 * 10 + 7 * 1.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q003. How would we write .00326 in terms of powers of 10?

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Your solution:

0 * .1 + 0 * .01 + 3 * .001 + 2 * .0001 + 6 * .00001 or 0 * 0 * 10^-1 + 0 * 10^-2 + 3 * 10^-3 + 2 * 10^-4 + 6 * 10^-5

confidence rating #$&*: 3

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Given Solution:

First we note that

.1 = 1/10 = 1/10^1 = 10^-1,

.01 = 1/100 = 1/10^2 = 10^-2,

.001 = 1/1000 = 1/10^3 = 10^-3, etc..

Thus .00326 means

0 * .1 + 0 * .01 + 3 * .001 + 2 * .0001 + 6 * .00001 =

0 * 10^-1 + 0 * 10^-2 + 3 * 10^-3 + 2 * 10^-4 + 6 * 10^-5 .

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q004. How would we add 3 * 10^2 + 5 * 10^1 + 7 * 10^0 to 5 * 10^2 + 4 * 10^1 + 2 * 10^0?

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Your solution:

First I wrote the problem out like this (3 * 10^2 + 5 * 10^1 + 7 * 10^0) + (5 * 10^2 + 4 * 10^1 + 2 * 10^0). Then I put the 10^2s, 10^1s, and 10^0s together like this: (3 * 10^2 + 5 * 10^2) + (5 * 10^1 + 4 * 10^1) + (7 * 10^0 + 2 * 10^0). That would equal 8 * 10^2 + 9 * 10^1 + 9 * 10^0, which equals 899.

confidence rating #$&*: 3

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Given Solution:

We would write the sum as

(3 * 10^2 + 5 * 10^1 + 7 * 10^0) + (5 * 10^2 + 4 * 10^1 + 2 * 10^0) ,

which we would then rearrange as

(3 * 10^2 + 5 * 10^2) + ( 5 * 10^1 + 4 * 10^1) + ( 7 * 10^0 + 2 * 10^0),

which gives us

8 * 10^2 + 9 * 10^1 + 9 * 10^0. This result would then be written as 899.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q005. How would we add 4 * 10^2 + 7 * 10^1 + 8 * 10^0 to 5 * 10^2 + 6 * 10^1 + 4 * 10^0?

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Your solution:

Like before, I rearranged to get (4 * 10^2 + 5 * 10^2) + ( 7 * 10^1 + 6 * 10^1) + ( 8 * 10^0 + 4 * 10^0). That equaled to 9 * 10^2 + 13 * 10^1 + 12 * 10^0. The 12 * 10^0 needs to be changed to 2 * 10^0 + 10^1. I added the 10^1 to 13 + 10^1 to get 14 + 10^1. Then did the same for 12 * 10^0 to get 4 * 10^1 + 10^2 and added the 10^2 to 9 * 10^2 to get 10 * 10^2. Then I did the same as before and ended up with 1 * 10^3 + 0 * 10^2 + 4 * 10^1 + 2 * 10^0, which equals 1042.

confidence rating #$&*: 3

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Given Solution:

We would write the sum as

(4 * 10^2 + 7 * 10^1 + 8 * 10^0) + (5 * 10^2 + 6 * 10^1 + 4 * 10^0) ,

which we would then rearrange as

(4 * 10^2 + 5 * 10^2) + ( 7 * 10^1 + 6 * 10^1) + ( 8 * 10^0 + 4 * 10^0),

which gives us

9 * 10^2 + 13 * 10^1 + 12 * 10^0.

Since 12 * 10^0 = (2 + 10 ) * 10^0 = 2 * 10^0 + 10^1, we have

9 * 10^2 + 13 * 10^1 + 1 * 10^1 + 2 * 10^0 =

9 * 10^2 + 14 * 10^1 + 2 * 10^0.

Since 14 * 10^1 = 10 * 10^1 + 4 * 10^1 = 10^2 + 4 * 10^1, we have

9 * 10^2 + 1 * 10^2 + 4 * 10^1 + 2 * 10^0 =

10^10^2 + 4 * 10^1 + 2 * 10^0.

Since 10*10^2 = 10^3, we rewrite this as 1 * 10^3 + 0 * 10^2 + 4 * 10^1 + 2 * 10^0.

This number would be expressed as 1042.

STUDENT SOLUTION

(4 x 10^2 + 5 x 10^2) + (7 x 10^1 + 6 + 10^1) + (8 x 10^0 + 4 x 10^0)

adds up to

9 x 10^2 + 13 x 10^1 + 12 x 10^0 = 1042

INSTRUCTOR RESPONSE

You got

9 x 10^2 + 13 x 10^1 + 12 x 10^0 = 1042

But this isn't in its final powers-of-10 notation.

13 * 10^1 isn't a legal expression. Since 13 is greater than 9, you would use the fact that 13 * 10^1 = 10^2 + 3 * 10^1 to write this in correct notation.

Your expression would then become

9 x 10^2 + 10^2 + 3 x 10^1 + 12 x 10^0

Also 12 * 10^0 = 10^1 + 2 * 10^0, so your expression is equivalent to

9 x 10^2 + 1 * 10^2 + 3 x 10^1 + 10^1 + 2 x 10^0

When we add the like powers of 10 we find that 9 * 10^2 + 10^2 = 10 * 10^2, which is 10^3.

Since 3 * 10^1 + 10^1 = 4 * 10^1.

your final expression should be

10^3 + 4 * 10^1 + 2 * 10^0.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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math41QA

&#Very good work. Let me know if you have questions. &#