The Biconditional statement
If two simple statements p and q are connected by the connective 'if and only if', then the resulting compound statement is called the biconditional statement. Symbolically it is represented by p q. Example: An integer is even if and only if it is divisibl..
If two simple statements p and q are connected by the connective 'if and only if', then the resulting compound statement is called the biconditional statement. Symbolically it is represented by p q. Example: An integer is even if and only if it is divisibl..Conditional and Biconditional Statements
Conditional and Biconditional Statements - It is already mentioned in earlier classes that compound statements of different propositions can be obtained by conjunction, disjunction and negation of propositions. We just recall these three basic logical connectivities an..
Momentary Lapse playing their original song "Bi-Conditional Statement" at the Health Tree in Palmdale, Ca. Nov. 17, 2007.
Access full lesson containing this video at: www.yourteacher.com Students learn that a conditional statement is an "if-then" statement. Students are then given conditional statements, and are asked to identify the hypothesis, the conclusion, the converse, the biconditional, and a counterexample (if applicable).
Question : I am trying to write out the negations of the definitions of upper and lower bounds and I cant remember how to write the negation of a biconditional statement.
The statements I have to negate are
Suppose S is a subset of Real numbers.
A number u in R is an upper bound if and only if for every s in S, s is less than or equal to u.
A number Z in R is a lower bound for s if and only if for every s in S, l is less than or equal to s.
Answer : P iff Q is equivalent to: (if P then Q) and (if Q then P). The negation of this is: not(if P then Q) or not(if Q then P). That last is equivalent, via contrapositives, to: (Q and notP) or (P and notQ)... More from Yahoo Answers
Answer : P iff Q is equivalent to: (if P then Q) and (if Q then P). The negation of this is: not(if P then Q) or not(if Q then P). That last is equivalent, via contrapositives, to: (Q and notP) or (P and notQ)... More from Yahoo Answers
Question : I have to prove the following by method of contradiction:
Let p and q be positive integers with p
Answer : a) The solution seems mysterious, until it smacks you in the face like a giant smacking thing (yes, I _do_ need to work on my analogies). Suppose (q-p) | (p-1). Then k Z such that k(q-p) = p-1. Adding q-p to both sides yields (k+1)(q-p) = q-1, so (q-p) | (q-1). Conversely, (q-p) | q-1 k Z s.t. k(q-p) = q-1 k Z s.t. (k-1)(q-p) = p-1 (q-p) | (p-1). b) Suppose d|(p-1) and d|(q-1). Then k , k s.t. k d = p-1 and k d = q-1, so (k - k )d = (q-1) - (p-1) = q-p, so d|(q-p). I'm not sure why your teacher asked you to use an indirect proof here -- as you can plainly see, the direct proofs are so much simpler. I wonder what she actually had in mind..... More from Yahoo Answers
Answer : a) The solution seems mysterious, until it smacks you in the face like a giant smacking thing (yes, I _do_ need to work on my analogies). Suppose (q-p) | (p-1). Then k Z such that k(q-p) = p-1. Adding q-p to both sides yields (k+1)(q-p) = q-1, so (q-p) | (q-1). Conversely, (q-p) | q-1 k Z s.t. k(q-p) = q-1 k Z s.t. (k-1)(q-p) = p-1 (q-p) | (p-1). b) Suppose d|(p-1) and d|(q-1). Then k , k s.t. k d = p-1 and k d = q-1, so (k - k )d = (q-1) - (p-1) = q-p, so d|(q-p). I'm not sure why your teacher asked you to use an indirect proof here -- as you can plainly see, the direct proofs are so much simpler. I wonder what she actually had in mind..... More from Yahoo Answers
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