Because it cannot be expressed in the form p/q where p and q are integers and q > 0.
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
The sum of p and q means (p+q). The difference of p and q means (p-q).
Not sure I can do a table here but: P True, Q True then P -> Q True P True, Q False then P -> Q False P False, Q True then P -> Q True P False, Q False then P -> Q True It is the same as not(P) OR Q
q + p
If p = 50 of q then q is 2% of p.
If p then q is represented as p -> q Negation of "if p then q" is represented as ~(p -> q)
p in the ss=planets in the solar system!
p-q
P! / q!(p-q)!
The assertion in the question is not always true. Multiplying (or dividing) 0 by a negative number does not yields 0, not a negative answer.Leaving that blunder aside, let p and q be positive numbers so that p*q is a positive number.Thenp*q + p*(-q) = p*[q + (-q)] = p*[q - q] = p*0 = 0that is p*q + p*(-q) = 0Thus p*(-q) is the additive opposite of p*q, and so, since p*q is positive, p*(-q) must be negative.A similar argument works for division.
By definition, every rational number x can be expressed as a ratio p/q where p and q are integers and q is not zero. Consider -p/q. Then by the properties of integers, -p is an integer and is the additive inverse of p. Therefore p + (-p) = 0Then p/q + (-p/q) = [p + (-p)] /q = 0/q.Also, -p/q is a ratio of two integers, with q non-zero and so -p/q is also a rational number. That is, -p/q is the additive inverse of x, expressed as a ratio.