Page:The Rhind Mathematical Papyrus, Volume I.pdf/78

The answer in this case might have been written^{[1] 1}⁄₂¹⁄₅, but the Egyptian thinks of ²⁄₃ as the largest fraction that there is and prefers to use it wherever he can.

In this problem and the next two the double of ²⁄₃¹⁄₁₀¹⁄₃₀ is given as ¹⁄₁¹⁄₃¹⁄₁₀. Directly it is 1¹⁄₃¹⁄₅¹⁄₁₅ Perhaps these fractions were taken as parts of some number, say 30. They would make 18, or 15 and 3, which would be ¹⁄₂ and ¹⁄₁₀ of 30.

Problem 5 Divide 8 loaves among 10 men.

Each man receives ²⁄₃¹⁄₁₀¹⁄₃₀.

Proof. Multiply ²⁄₃¹⁄₁₀¹⁄₃₀ by 10; the result is 8.
Do it thus:

	1	2⁄31⁄101⁄30
\	2	11⁄21⁄10
&nsbp;	4	31⁄5
\	8	61⁄31⁄15

Total 8 loaves, which is correct.

Problem 6
Divide 9 loaves among 10 men.

Each man receives ²⁄₃¹⁄₅¹⁄₃₀.

Proof. Multiply ²⁄₃¹⁄₅¹⁄₃₀ by 10.
Do it thus:

	1	2⁄31⁄51⁄30
\	2	12⁄31⁄101⁄30
	4	31⁄21⁄10
\	8	71⁄5

Total 9 loaves, which is correct.

In the first doubling we have 1¹⁄₃,¹⁄₃¹⁄₁₅, and ¹⁄₁₅. The two ¹⁄₃'s make ²⁄₃ and the two ¹⁄₁₅'s make ¹⁄₁₀¹⁄₃₀. Thus we get 1²⁄₃¹⁄₁₀¹⁄₃₀. The next doubling is the same as in Problem 4.

SECTION III

Problems 7-20. Multiplication by Certain Fractional Expressions^[2]

Problem 7
Multiply ¹⁄₄¹⁄₂₈ by 1¹⁄₂¹⁄₄.

1

1⁄4

1⁄28

as

parts

of

28

these

are

7

and

1

1⁄2

1⁄8

1⁄36

"

"

"

"

"

"

31⁄2

"

1⁄2

1⁄4

1⁄16

1⁄112

"

"

"

"

"

"

11⁄21⁄4

"

1⁄4

Total

1⁄2 since as a part of 28 this is 14.

1. Problem 54 is numerically the same problem, but the division is carried through there in such a way that ¹⁄₂¹⁄₅ is the form of the result.
2. See Introduction, page 23.