p divided by q.
If ( p ) is an integer and ( q ) is a nonzero integer, then the expression ( \frac{p}{q} ) will always yield a rational number. Additionally, since ( q ) is nonzero, ( p ) cannot be divided by zero, ensuring the division is valid. Furthermore, ( p + q ) will also be an integer, as the sum of two integers is always an integer.
The sum of p and q means (p+q). The difference of p and q means (p-q).
The expression ( p \land q ) is called the conjunction of ( p ) and ( q ). It represents the logical operation where the result is true only if both ( p ) and ( q ) are true. If either ( p ) or ( q ) is false, the conjunction ( p \land q ) is false.
No, the statement "not(p and q)" is not equal to "(not p) or q." According to De Morgan's laws, "not(p and q)" is equivalent to "not p or not q." This means that if either p is false or q is false (or both), the expression "not(p and q)" will be true. Therefore, the two expressions represent different logical conditions.
q + p
p divided by q.
3/p = 6 p = 3/6 = 1/2 3/q = 15 q = 3/15 = 1/5 p - q = 1/2 - 1/5 p - q = 5/10-2/10 = 3/10
Undefined: You cannot divide by zero
I'm on it . . .p = 2 / (m + q)Multiply each side by (m + q) :p (m + q) = 2Divide each side by 'p' :m + q = 2/pSubtract 'm' from each side:q = 2/p - m
Any rational number (by definition).
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
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-q