If we multiply the number 1 by a unit A, we end up with one unit of A:
1 x A = A
We are at liberty to substitute A with any quantity we wish:
1 x 5kg = 5kg
1 x 6s = 6s
1 x 100m = 100m
Furthermore,
1 x A = A x 1
1 x 5kg = 5kg x 1
1 x 6s = 6s x 1
1 x 100m = 100m x 1
The expressions above show how the “ commutative properties “ of algebra allow us to rearrange terms without altering the identity of the expression at hand. This mathematical principle enables us to take known quantities and express them in new terms that we choose; however, it will be useful to demonstrate how the number 1 can be derived from ratios of our choosing as well:
( A / A ) = 1
( 5kg / 5kg ) = 1
( 6s / 6s ) = 1
( 100m / 100m ) = 1
To further expand things a bit, let’s look at how the commutative properties of fractions will aid us in our efforts. Brackets ( ) are used to enclose fractions that are being multiplied by one another:
( 1 /2 )( 2 / 1 ) = ( 2 / 2 )( 1 / 1 ) = 1
( 1 / 4 )( 3 / 4 ) = ( 3 / 4 )( 1 / 4 ) = ( 3 / 16 )
( a / b )( c / d ) = ( c / b )( a / d ) = [ ( ca ) / ( bd ) ]
( abc / def )( hij / xyz ) = ( hij / def )( abc / xyz ) = [ ( hij )( abc ) / ( def )( xyz ) ]
In an example below, we will be given two choices of conversion fractions to be used to express 5 seconds ( 5s ) in terms of milliseconds ( ms ). We will derive our conversion fraction from the following true statement:
1ms = 10-3 s
ANY EQUIVALENT STATEMENT CAN BE USED TO FORM THE NUMBER 1 IN THE FORM OF A FRACTION. The equivalency above can be used to form the following two fractions whose values are equal to the number 1:
a. ( 1ms / 10-3 s ) = 1
b. ( 10-3 s / 1ms ) = 1
Here’s another way of looking at things:
a. ( 1 / 10-3 )( ms / s ) = 1
Since 1ms = 10-3 s, we are at liberty to use substitution to simplify the expression:
a. ( 1 / 10-3 )( 10-3 s / s ) = 1
a. ( 1 / 10-3 )( s / s )( 10-3 s / 1 ) = 1
a. ( s / s )( 1 / 1 )( 10-3 / 10-3 ) = 1
Note: It wasn’t necessary to write the number 1 in the second expression ( 10-3 s / 1 ) above, because it is understood that any quantity divided by 1 is equal to the quantity at hand:
Ex. ( 5s / 1 ) = 5s
Ex. ( 10-3 s / 1 ) = 10-3 s
Let’s now convert 5s to an equivalent statement expressed in terms of milliseconds. Recall that the number 1 was formed twice using the same statement of truth:
1ms = 10-3 s
Which of the expressions below can be used to correctly convert 5s to its millisecond equivalent?
a. ( 5s )( 1ms / 10-3 s )
b. ( 5s )( 10-3 s / 1 ms )
The commutative property can be used to let us know which fraction to use. Let’s first examine ( b ):
b. ( 5s )( 10-3 s / 1ms ) = ( 5s / 1ms )( 10-3 s ) = ( 5s2 )( 10-3 / 1ms)
No conversion occurred! The final expression is different in composition compared to the one from which it was derived. Thus, the expression that contains the ( 1ms / 10-3 s ) term is correct:
a. ( 5s )( 1ms / 10-3 s ) = ( 5s / s )( 1ms / 10-3 ) = ( 5 )( 1ms / 10-3 )
a. ( 5 / 10-3 )( 1ms ) = ( 5,000 )( 1ms ) = 5,000ms
5s = 5,000ms
THE “ ZIG ZAG “ APPROACH
The following method can be used to further practice using conversion fractions in their correct format:
( 1kg )( 103 g / 1kg )( 1mg / 10-3 g ) = ???
Moving from left to right, notice how the units in the numerator and denominator match up in a “ zig zag “ type fashion. The expression begins with a unit of ( 1kg ). The unit of ( kg ) is contained in the denominator of the next expression. Even though ( 103 g ) and ( 10-3 g ) multiply ( g ) by different multiplicands, both expressions are written in units of grams ( g ). This makes the setup correct:
( 1kg )( 103 g / 1kg )( 1mg / 10-3 g ) = ???
( 1kg / 1kg )( 103 g / 10-3 g )( 1mg ) = ???
( 1 )( 106 )( 1mg ) = ???
??? = 106 mg
A kilogram of any substance contains 1 million milligrams.
Check: ( 106 )( 10-3 )( g ) = ( 103 )( g ) = 1kg
RENAISSANCE PHYSICS 