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Why is the sum of two rational numbers always rational numbers?
Suppose p/q and r/s are rational numbers where p, q, r and s are integers and q, s are non-zero.Then p/q + r/s = ps/qs + qr/qs = (ps + qr)/qs. Since p, q, r, s are integers, then ps and qr are integers, and therefore (ps + qr) is an integer. q and s are non-zero integers and so qs is a non-zero integer. Consequently, (ps + qr)/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.
Why is the sum or product of two rational numbers rational?
Suppose p/q and r/s are rational numbers where p, q, r and s are integers and q, s are non-zero.Then p/q + r/s = ps/qs + qr/qs = (ps + qr)/qs.Since p, q, r, s are integers, then ps and qr are integers, and therefore (ps + qr) is an integer.q and s are non-zero integers and so qs is a non-zero integer.Consequently, (ps + qr)/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.Also p/q * r/s = pr/qs.Since p, q, r, s are integers, then pr and qs are integers.q and s are non-zero integers so qs is a non-zero integer.Consequently, pr/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.
Why is the difference of two rational numbers are rational numbers?
Suppose A and B are two rational numbers. So A = p/q where p and q are integers and q > 0 and B = r/s where r and s are integers and s > 0. Then A - B = p/q - r/s = ps/qs - qr/qs = (ps - qr)/qs Now, p,q,r,s are integers so ps and qr are integers and so x = ps-qr is an integer and y = qs is an integer which is > 0 Thus A-B can be written as a ratio of two integers, x/y where y>0. Therefore, A-B is rational.
Why is the difference between two rational numbers always a rational number?
Suppose x and y are two rational numbers. Therefore x = p/q and y = r/s where p, q, r and s are integers and q and s are not zero.Then x - y = p/q - r/s = ps/qs - qr/qs = (ps - qr)/qsBy the closure of the set of integers under multiplication, ps, qr and qs are all integers,by the closure of the set of integers under subtraction, (ps - qr) is an integer,and by the multiplicative properties of 0, qs is non zero.Therefore (ps - qr)/qs satisfies the requirements of a rational number.
What is the plural possessive of p and q?
The plural form is Ps and Qs.The plural possessive form is Ps and Qs'.Example: Your Ps and Qs' training seems lacking.
How you can add two rational numbers?
If the two rational numbers are expressed as p/q and r/s, then their sum is (ps + rq)/(qs)
How do you subtract rational numbers with the same or different signs?
Suppose x and y are two rational number.Then x = p/q and y = r/s where p, q, r and s are integers, with q and s being non-zero. Then x - y = p/q - r/s = pq/qs - qr/qs = (pq - rs)/qs. The signs of x and y do not matter, in so far as their signs will be used to determine the signs of p,q, r and s.
Why are rational numbers closed under addition?
Suppose x and y are rational numbers.That is, x = p/q and y = r/s where p, q, r and s are integers and q, s are non-zero.Then x + y = ps/qs + qr/qs = (ps + qr)/qsThe set of integers is closed under multiplication so ps, qr and qs are integers;then, since the set of integers is closed addition, ps + qr is an integer;and q, s are non-zero so qs is not zero.So x + y can be represented by a ratio of two integers, ps + qr and qs where the latter is non-zero.
What does the abbreviation q c stand for?
Queen's Counsel
Words with double q?
None in English, if you mean paired Qs.
Is the dfference of two positive rational number always positive plz help explain.?
Yes.Suppose a and b are two positive rational numbers. Then a can be expressed in the form p/q where p and q are positive integers, and b can be expressed in the form r/s where r and s are positive integers.Then b - a = r/s - p/q = (qr - ps)/qs.Now, since p, q, r and s are integers, thenby the closure of the set of integers under multiplications, qr, ps and qs are integers;q and s are positive => qs is positive, andby the closure of the set of integers under addition (and subtraction), qr - ps is an integer.That is, b - a = (qr - ps)/qs is a ratio of two integers, where the denominator of the ratio is positive.
What does q c stand for in the highway q c's?
Hey, You should try the link given below: (related to programming) http://q-lang.sourceforge.net/qcalc/qdoc_15.html (If it is QCC it can be Quality Control Circle also or Quality Core Curriculum)
