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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 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.
How do you divide multiply fraction?
You do it the obvious way. Take p, q in Q, the rationals By definition, we can write p = m/n and q = r/s where m, r are integers, n and s are natural. we define pq (p times q) = (mr)/(ns) p/q = pq^-1 where q^-1 denotes q's multiplicative inverse s/r Remark: you cannot divide by 0 here because 1) 0 have no multiplicative inverse 2) if r = 0. s/r is undefined.
What is the geometrical proof of a-b whole square?
Given the graphic capability of this site, you are going to have to use some imagination! <---------a---------> <---a-b---><--b--> +-----------+-------+ |...............|..........| |.......P......|....Q...| |...............|..........| +-----------+-------+ |.......R......|....S....| |...............|..........| +-----------+-------+ In the above graphic, P, S and the whole figure are meant to be squares. The total area is P+Q+R+S = a2 P = (a-b)2 Q = b*(a-b) = (a-b)*b = a*b - b2 R = (a-b)*b = a*b = a*b - b2 and S = b2 Now, P = {P+Q+R+S} - Q - R - S = a2 - ab + b2 - ab + b2 - b2 = a2 - 2ab + b2
P varies jointly as q r and s One set of values is p equals 70 q equals 7 r equals 5 and s equals 4 Find p when q equals 2 r equals 15 and s equals 7?
If P varies jointly as q, r and s - assume this is in direct proportion, then P ∝ qrs so P = kqrs where k is a constant.70 = k x 7 x 5 x 4 = 140k : k = 140/70 = 0.5When q = 2, r = 15 and s = 7 then,P = 0.5 x 2 x 15 x 7 = 105
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)
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.
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.
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.
Can a 3 X 4 matrix be multiplied by a 4 X 1 matrix?
Yes. If one matrix is p*q and another is r*s then they can be multiplied if and only if q = r and, in that case, the result is a p*s matrix.
