6
RECIPROCALS
AND
ZERO
The quotient of a divided by b
TWO NUMBERS ARE CALLED reciprocals of one another when their product is 1.
| The reciprocal of 2, for example, is | 1 2 |
-- because |
| 2· | 1 2 |
= 1. |
| a· | 1 a |
= | 1 a |
· a | = 1 |
The student should not do arithmetic -- that is, "cancel" the a's. The
| student should recognize by the written form itself that | a· | 1 a |
= 1. |
Problem 1. Write the symbol for the reciprocal of z.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
|
1 z |
| Problem 2. If pq = 1, then q = | 1 p |
. q is the reciprocal of p. |
| Problem 3. What is the meaning of the symbol | 1 log 2 |
? |
It is that number which, when multiplied with log 2, produces 1.
And it is not even necessary to know what "log 2" means
The statement
| 1 log 2 |
· log 2 = 1 |
is purely formal.
| Problem 4. Prove that the reciprocal of | 3 4 |
is | 4 3 |
. |
| 3 4 |
· | 4 3 |
= 1 |
To "prove" anything, we must satisfy its definition. In this case, the definition of reciprocals is that their product is 1.
| In general, the reciprocal of a fraction | a b |
is | b a |
. |
Problem 5.
| a) |  |
= | 2. | 1 over any number is its reciprocal. |
| b) |  |
= | 3 2 |
c) | x ¼ab |
= | x ab 4 |
= | 4x ab |
Problem 6.
| a) | xyz· | 1 xyz |
= | 1 | b) | (a − b) | · | 1 a − b |
= 1 |
| c) | (p + q)· | 1 (p + q) |
= 1 | d) | &8%· | 1 &8% |
= | 1 |
The definition of division
We say, in algebra, that division is multiplication by the reciprocal.
| a b |
= a· | 1 b |
= | 1 b |
· a |
"a divided by b is equal to a times the reciprocal of b."
Equivalently,
We may rewrite any fraction as the numerator
times the reciprocal of the denominator.
That rule in the box is called The Definition of Division. Division is stated in terms of multiplication, just as subtraction can be stated in terms of addition: a − b = a + (−b). (Lesson 3.)
Problem 7. Rewrite each of the following according to The Definition of Division.
| a) | 3 4 |
= | 3· | 1 4 |
b) | x 2 |
= | x· | 1 2 |
c) | x + 1 x + 2 |
= | (x + 1)· | 1 (x + 2) |
| d) | a + b + c 6 |
= | (a + b + c)· | 1 6 |
| Problem 8. Prove that | a a |
= 1 , for any number a 0. |
| a a |
= | a· | 1 a |
, according to the Definition of Division, |
| = | 1, | according to the definition of reciprocals. | ||
In other words, when the numerator and denominator are equal, the fraction is immediately equal to 1. This has nothing to do with "canceling."
Problem 9. Evaluate each of the following. (Assume that no denominator is 0.)
| a) | x − 2 x − 2 |
= | 1 | b) | a + b + c a + b + c |
= | 1 | c) | −(x + 5) −(x + 5) |
= | 1 |
| d) | x −x |
= | −1 | e) | a + b −(a + b) |
= | −1 | f) | −(x²
+ 5x − 2) x² + 5x − 2 |
= | −1 |
Inverse operations
Division and multiplication are inverse operations. That means that if we start with some number x, and then divide it by a number a, then to preserve x we have to multiply by a:
| x a |
· | a | = | x |
Or, if we first multiply by a, then to preserve x we have to divide
| x· a· | 1 a |
= | x |
| by a, or, equivalently, multiply by | 1 a |
. |
| 1 a |
is also called the multiplicative inverse of a, |
| because when we multiply it with a, it produces 1, which is the identity. | |
| Compare Lesson 10 of Arithmetic: Property 1 of division. | |
Problem 10. Complete each equality with some operation, so that the equality is preserved.
| a) 27 = 27· 3 | · | 1 3 |
| b) 27 = | 27 3 |
· | 3 |
| c) p = | p q |
· | q |
| d) m = mn | · | 1 n |
| e) cos x = | cos x sin x |
· | sin x |
The student may think that is trigonometry, but it is not. It is algebra.
| f) cos x = 2 cos x sin x | · |
__1__ 2 sin x |
| The quotient of | a b |
The rule called the Definition of Division --
| a b |
= a· | 1 b |
| -- is merely a formal rule. It tells us that we may replace | 12 3 |
with 12· | 1 3 |
. |
But it does not tell us how to evaluate that division. For that, we must return to arithmetic, and to the relationship between division and multiplication.
If the quotient of a divided by b is some number n --
| a b |
= | n |
-- then n is that number such that n times b is equal to a.
a = nb.
For example,
| 12 3 |
= | 4, |
because
4· 3 = 12.
We will be applying this below.
Rules for 0
a· 0 = 0· a = 0
"If any factor is 0, the product will be 0."
Problem 11.
| a) | 9· 0 = 0 | b) | 0· 9 = 0 | c) | 7· 45· 127· 0· 39 = 0 |
Problem 12. If the product of two factors is 0,
ab = 0,
what can you conclude about a or b?
Either a = 0 or b = 0.
Problem 13. Which values of x will make this product equal 0?
(x − 1)(x + 2)(x + 3) = 0
Each value of x that will make each factor equal to 0.
In the first factor, x = 1.
In the middle factor, x = −2.
And in the last factor, x = −3.
For those three values -- and only those three -- will that product equal 0.
*
We will now investigate the following forms. In each one, a
0.
| 0 a |
= ? |
| a 0 |
= ? |
| 0 0 |
= ? |
Problem 14.
| a) | 8 2 |
= n. What number is n? |
n = 4. Because according to the meaning of the quotient n, 4· 2 = 8.
| b) | 0 8 |
= n. What number is n? |
n = 0. Because 0· 8 = 0.
| Therefore, | 0 a |
= 0, for any number a 0. |
| c) | 8 0 |
= n. What number is n? |
There is no number n such that n· 0 = 8. Division by 0
| is an excluded operation. The symbol | 8 0 |
-- although it |
| may look like a number -- is not a number. | 8 0 |
is a |
meaningless symbol. It has no value.
(The student should not confuse no value with the value 0. 0 is a perfectly good number.)
| d) | 0 0 |
= n. What number is n? |
n could be any number, because any number n· 0 = 0. The
| symbol | 0 0 |
could have any value. It is truly "undefined." |
In summary (a0):
| 0 a |
= 0 |
| a 0 |
= No value. Meaningless. |
| 0 0 |
= Any value. |
(For a "proof" that 1 + 1 = 1, click here.)
| It is common to hear that | 8 0 |
is "undefined." Rather, |
| it is undefinable. Elsewhere in mathematics, for example, we say that this quotient | ||
| 5x x |
is "undefined" for x = 0. Nevertheless, since
| 5x x |
= 5 |
for all other values of x, we can define that quotient as 5 -- or any number we please -- when x = 0. And
| there is no problem. For, | 0 0 |
may be any number. | |||||||||||
| However we can never define the symbol | 8 0 |
. |
Problem 15.
| a) | 0 5 |
= | 0 | b) | 0 −5 |
= | 0 | c) | 0 x |
(x0) = |
0 | ||
| d) | 5 0 |
= | No value. | e) | 0 0 |
= | Any value. | f) | x 0 |
(x0) = |
No value. | ||
Problem 16. Let x = 2, and evaluate the following.
| a) | x − 2 x + 2 |
= | 0 4 |
= 0 | b) | x + 2 x − 2 |
= | 4 0 |
= No value. |
| c) | 2x − 4 3x − 6 |
= | 0 0 |
= Any value. |
Problem 17. Does 0 have a reciprocal?
| No. There is no number | 1 0 |
. |
Problem 18. What is the only way that a fraction could be equal to 0?
The numerator must be equal to 0.
Problem 19. Solve for x.
| a) | x − 2 x + 5 |
= 0 | b) | x + 3 x − 1 |
= 0 | ||
| x = 2 | x = −3 | ||||||
See the previous problem.
Next Lesson: Removing grouping symbols
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