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How do you add and subtract rational numbers?
A rational number is a number of the form p/q where p and q are integers and q > 0.If p/q and r/s are two rational numbers thenp/q + r/s = (p*s + q*r) / (q*r)andp/q - r/s = (p*s - q*r) / (q*r)The answers may need simplification.
Prove that if a and b are rational numbers then a multiplied by b is a rational number?
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
What are the rules for multiplying rational numbers?
p/q * r/s = (p*r)/(q*s)
What is the proof of transitivity in logic NOT with truth table if p is q and if r is s and either p or r is true therefore either q or s is true?
Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
L is not as tall as P or R but is taller than S and Q Q being shorter than R S is shorter than R who is shorter than P who is taller than Q Who among P Q R and S is the shortest?
The answer is Q.
What is the answer to add q and p triple the result then divide r by s?
It is 3*(q + p)/(r + s)
What do you do while multiplying rational numbers?
A rational number is a number which can be expressed in the form p/q where p and q are integers and p>0.If p/q and r/s are two rational numbers then(p/q)*(r/s) = (p*r)/(q*s).You may need to check that this fraction is in its lowest (simplest) form.
Can the product of two rational numbers be irrational?
No.Suppose a and b are two rational numbers.Then they can be written as follows: a = p/q, b = r/s where p, q, r and s are integers and q, s >0.Then a*b = (p*r)/(q*s).Using the properties of integers, p*r and q*s are integers and q*s is non-zero. So a*b can be expressed as a ratio of two integers and so the product is rational.
How do you use the LCM to write fractions with a common denominator?
Suppose you have the fractions p/q and r/s. Let the LCM of q and s be t.Then t is a multiple of q as well as of s so let t= q*u and t = s*v Then p/q = (p*u)/(q*u) = (p*u)/t and r/s = (r*v)/(s*v) = (r*v)/t have the same denominators.
How do you rename as improper faction As a mixed number?
Suppose the improper fraction is p/q, then if q goes into p r times with a remainder of s, thenp/q = r s/q
An equation that shows two ratios are equivalent?
Two ratios, p/q and r/s (q and s non-zero) are equal if p/q - r/s = 0.
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.
