Lesson 2, Topic 3
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# Binary Arithmetic Operations – Addition | Subtraction in Other Bases

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a. 3042five   + 2103five

Explanation:

1st Column (from right to left) = 2 + 3 = 5. How many fives are there in 5? $$\frac{5}{5}$$= 1 remainder 0. Write 0 and carry 1 to column 2.

2nd Column = 1 + 4 + 0 = 5. How many fives are there in 5? $$\frac{5}{5}$$= 1 remainder 0. Write 0 and carry 1 to column 3.

3rd Column = 1 + 0 + 1 = 2. 2 is less than 5 so we write 2 down.

4th column = 3 + 2 = 5. How many fives are there in 5? $$\frac{5}{5}$$= 1 remainder 0. Write 0 and carry 1 to column 5.

5th column = We add 1 to the 5th column.

b. 2001three   + 1022three

Explanation

1st Column (from right to left) = 1 + 2 = 3. How many threes are there in 3? $$\frac{3}{3}$$= 1 remainder 0. Write 0 and carry 1 to column 2.

2nd Column = 1 + 0 + 2 = 3. How many threes are there in 3? $$\frac{3}{3}$$= 1 remainder 0. Write 0 and carry 1 to column 3.

3rd Column = 1 + 0 + 0 = 1. 1 is less than 3 so we write 1 down.

4th column = 2 + 1 = 3. How many threes are there in 3? $$\frac{3}{3}$$= 1 remainder 0. Write 0 and carry 1 to column 5.

5th column = We add 1 to the 5th column.

c. 151.24six +   12.34six  + 20.11six

Explanation

1st Column (from right to left) = 4 + 4 + 1 = 9. How many six’s are there in 9? $$\frac{9}{6}$$ = 1 remainder 3. Write 3 and carry 1 to column 2.

2nd Column = 1 + 2 + 3 + 1 = 7. How many six’s are there in 7? $$\frac{7}{6}$$ = 1 remainder 1. Write 1 and carry 1 to column 3.

3rd Column = 1 + 1 + 2 + 0 = 4. 4 is less than 6 so we write 4 down.

4th column = 5 + 1 + 2 = 8. How many six’s are there in 8? $$\frac{8}{6}$$ = 1 remainder 2. Write 2 and carry 1 to column 5.

5th column = 1 + 1 = 2. 2 is less than 6 so we write 6 down.

d. 53468 – 6378

1st Column (from right to left) = 6 is less than 7 so we will have to borrow from the 2nd column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the second column will give (8 + 6) – 7 = 14 – 7 = 7

2nd Column = This reduces 4 to 3 in the second column. 3 – 3 = 0. We write 0 down.

3rd Column = 3 is less than 6 so we will have to borrow from the 4th column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the fourth column will give (8 + 3) – 6 = 11 – 6 = 5

4th Column = This reduces 5 to 4 in the fourth column. 4 – 0 = 4. We write 4 down.

e. 3002four – 2213four

Explanation

1st Column (from right to left) = 2 is less than 3 so we will have to borrow from the 2nd column. Since it is in base 4 any number borrowed is 4. So borrowing 4 from the second column will give (4 + 2) – 3 = 6 – 3 = 3

2nd Column = This reduces 0 to -1 in the second column. We will have to borrow (4) from the third column. 4 – 1 – 1 = 2. We write 2 down.

3rd Column = This reduces 0 to -1 in the third column. We will have to borrow (4) from the 4th column. 4 – 1 – 2 = 1. We write down 1.

4th Column = This reduces 3 to 2 in the fourth column. 2 – 2 = 0.