Consider the set of Rational numbers. {(a,b) s.t. a/b and a,b are integers with
ID: 2944080 • Letter: C
Question
Consider the set of Rational numbers.{(a,b) s.t. a/b and a,b are integers with b not 0}.
Let ~ be the usual equivalence relation that partitions the set Q of rational numbers.
For example, (1,2) ~ (2,4) because 2/4 reduces to 1/2.
show the following:
If (a,b)~(a',b') and (c,d)~(c',d'), then
(ad+bc,bd)~(a'd'+b'c',b'd').
And hence showing that + is a well defined operation on Q.
I will start out the proof to save you some time . . .
Suppose (a,b)~(a',b') and (c,d)~(c',d').
That's as far as I can get on this problem.
Explanation / Answer
Consider the set of Rational numbers.
{(a,b) s.t. a/b and a,b are integers with b not 0}.
Let ~ be the usual equivalence relation that partitions the set Q of rational numbers.
For example, (1,2) ~ (2,4) because 2/4 reduces to 1/2.
show the following:
If (a,b)~(a',b') and (c,d)~(c',d'), then (ad+bc,bd)~(a'd'+b'c',b'd')
And hence showing that + is a well defined operation on Q.
I will start out the proof to save you some time . . .
Suppose (a,b)~(a',b') and (c,d)~(c',d')
Thus a is the usual equivalence relation to a' and b is the usual equivalence relation to b'.
Thus a is a factor of a' and b is a factor of b'.
Since this is a set of Rational numbers, the commutative property of multiplication must apply. Since a ~ a' and d ~ d', we see that:
(a)(d) = ad and thus ad ~ a'd'
Similarly,
(b)(c) = bc and thus bc ~ b'c'
Similarly,
(b)(d) = bd and thus bd ~ b'd'.
Since this is the set of Rational numbers, the associative property of addition must apply.
Thus if ad ~ a'd' and bc ~ b'c', then ad + bc ~ a'd' + b'c'
Thus bd ~ b'd'.
We have thus proved all components of
If (a,b)~(a',b') and (c,d)~(c',d'), then (ad+bc,bd)~(a'd'+b'c',b'd')
and therefore proved that addition is a well-defined operation on Q.