SOLUTION: (a) Prove that the set S of rational numbers (in lowest term) with odd denominators is
a subring of Q.
(b) Let I be the set of elements of S with even numerators. Prove that I is
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Question 32844: (a) Prove that the set S of rational numbers (in lowest term) with odd denominators is
a subring of Q.
(b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in
S.
(c) Show that S/I consists of exactly two elements.
Thank You
Found 2 solutions by khwang, venugopalramana:
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
Prove that the set S of rational numbers (in lowest term) with odd denominators is
a subring of Q.
(b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in S.
(c) Show that S/I consists of exactly two elements.
proof: a) S = { p/q| p, q in Z,(p,q) = 1 & q is odd}
p1/q1, p2/q2 in S --> p1/q1- p2/q2 = (p1q2 - p2q1)/q1 q2 is in S
and p1/q1* p2/q2 = p1p2/q1q2 is in S
So, S is a subring of Q.
b) I = {p/q in S| p even}
As in a) test I is a subring of S (for you).
for r in S, p/q in I, so p is even and so
r * pq = pp'/qq' if r = p'/q' is in I since pp' is even.
Hence, rI < I and so I is an ideof Q.
c) S/I ={ s + I | s in S}
for any s in S, let s = p/q
s is in I if p is even.(i.e s+I = I)
if p is odd, then s - 1/3 = p/q - 1/3 = (3p-q)/3q belonging to I
(since 3p-q is even)
Hence, s + I = 1/3 for any s = p/q (p odd) in I.
This shows S/I = {I, 1/3 + I} with two elements.
Kenny
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
S IS THE SET OF RATIONAL NUMBERS WITH D.R. BEING ODD NUMBER ..THAT
IS ELEMENTS ARE OF THE TYPE P/(2Q-1)
TST S IS A SUB RING OF Q.
FOR THIS WE NEED TO SHOW
1.IF A AND B ARE ELEMENTS OF S THEN A-B IS ALSO AN ELEMENT OF S.
LET A=P1/(2Q1-1)...AND....B=P2/(2Q2-1)
A-B={P1(2Q2-1)-P2(2Q1-1)}/(2Q1-1)(2Q2-1)...WE FIND THAT D.R IS PRODUCT
OF 2 ODD NUMBERS AND HENCE IS ODD.SO THE RESULTANT VALUE OF A-B IS
STILL IN THE FORM OF SOME P/(2Q-1)...SO THIS BELONGS TO S
2.IF A AND B ARE ELEMENTS OF S THEN AB IS ALSO AN ELEMENT OF S
WITH SAME A AND B AS ABOVE WE HAVE
AB=(P1P2)/(2Q1-1)(2Q2-1)...WHICH IS AGAIN OF THE TYPE SOME P/(2Q-1)
AND HENCE AN ELEMENT OF S
THESE 2 CRITERIA BEING NECESSARY AND SUFFICIENT FOR S TO BE A SUB RING
OF Q WE CONCLUDE THE RESULT.
----------------------------------------------------------------------------------------------------------------------
I IS SET OF EVEN N.R....SO THE ELEMENTS ARE 2P/(2Q-1)..TYPE.
TO SHOW THAT THIS IS AN IDEAL OF S.
FOR THIS WE HAVE TO PROVE THAT
1.IF A AND B ARE ELEMENTS OF I ,THEN A-B IS AN ELEMENT OF I.
LET A =2P1/(2Q-1)....B=2P2/(2Q2-1)....
A-B = 2[P1(2Q2-1)-P2(2Q1-1)]/(2Q1-1)(2Q2-1).....WE FIND THAT N.R IS A
MULTIPLE OF 2 AND HENCE AN EVEN NUMBER.DENOMINATOR IS PRODUCT OF 2 ODD
NUMBERS AND HENCE AN ODD NUMBER.
HENCE A-B IS EQUAL TO SOME 2P/(2Q-1)...THAT IS IT BELONGS TO I
2.IF A IS AN ELEMENT OF I AND R IS AN ELEMENT OF S THEN A*R AND R*A
ARE ELEMENTS OF I.
LET R=R1/(2Q3-1)....WE HAVE
A*R=R*A=[(R1)(2P1)]/(2Q1-1)(2Q3-1)...WHICH IS AGAIN OF THE TYPE SOME
2P/(2Q-1)..FOR THE SAME REASONS GIVEN ABOVE.
THESE 2 CRITERIA BEING NECESSARY AND SUFFICIENT CRITERIA FOR I TO BE
AN IDEAL OF S ,WE CONCLUDE THE RESULT.
BY DEFINITION S/I IS THE SET OF COSETS OF I IN S
I/S={S+A:A IS AN ELEMENT OF S}
NOW WE HAVE I AS THE SET OF ELEMENTS OF THE TYPE...EVEN /ODD NUMBER
WHERE AS S IS THE SET OF ELEMENTS OF THE TYPE,....EVEN/ODD AND ODD/ODD
ELEMENTS...SO WHEN YOU MAKE A COSET , YOU WILL HAVE ONLY 2 DISTINCT
CLASSES ALWAYS NAMELY ODD AND EVEN IN NR.SINCE S/I IS THE SET OF
DISTINCT COSETS ,WE HAVE ONLY 2 DISTINCT CLASSES OR ELEMENTS IN THIS
QUOTIENT RING OR RESIDUAL CLASS AS IT IS CALLED.
HENCE THERE ARE ONLY 2 DISTINCT ELEMENTS IN S/I,ONE CLASS OF ELEMENTS OS 2P/(2Q-1)AND ANOTHER CLASS OF ELEMENTS OF TYPE (2P-1)/(2Q-1)
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