week 2 Math230
this truth table I made shows where the argument is invalid and valid
when p, q, is true and r is false . the argument is not true.
this truth table I made shows where the argument is invalid and valid.
look at when p is true and q is false. the argument is suggesting that when variable q and p is true the argument is true, however when q is false this is cant be the case. leading to our argument being invalid.
when p is true and q is false our argument is invalid.
if p was to study and q was successful
it is true , if you study you will be successful ok
so when you study you don't become successful is false
when you don't study you are successful is true ( the ARGUEMENT can still be true maybe you have a hardworking job ! )
if you don't study you are not successful is true ( you did not study and you are not successful)
this truth table I made shows where the argument is invalid and valid
when p is false and q is true the argument is invalid.
this implies it is not false and not true
its not false if you study and its not true you will be successful
which is saying its true to study and not be successful which is not true according to the other logic in the table.
the question asks us which of the following statements is valid or invalid. we will solve this by assigning each proposition with a variable
part
I will be assigning
x as the patient
k to high blood pressure
y to diabetes
m to high cholesterol
first statement is the hypothesis
second statement is incorrect because the proof wants to know k or y
not y or m
third statement is incorrect as well it has the k but not the y
this was an incorrect way to use the Resolution law if this is what this was trying to do.
Part
i would say both statements are math logically true to everything i learned in zylabs.
the first statement mentions a student and John is a student. It mentions a student got an A. Then it mentions John got an A. So this is logically correct. This is a form of Existential generalizatiom as the element was defined with the variable
the second statement mentions all girls scout who sold at least
Part
statement A is invalid because the and operator.
logically both sides need to be true and there is instances for when P is false or Q that make it invalid
Statement B is is valid because when the operator is an or symbol it doesn't matter whether the q or p is true or false it will be true. as long as both are not false we are good and the proof is valid. here is a proof by direct proof.
testing T and F when
problem
if abc are real integers, such that b =
then c = a * a * a * a * a * a
i will use
we find a
Problem
a^2 + b^2 + c^2 + d^2 =
is true when testing with
Problem
if the problem states x is rational prove that if xy is irrational than y is irrational
i will use contrapositive to prove this easy question.
rational numbers cannot be irrational numbers this is fundemental to math.
so when applying contrapositive logic to y.
we have y is a rational number such that xy is irrational. which makes this invalid
"An irrational number is a real number that is not rational. Note that the definition implies that every real number is either rational or irrational but not both." zylabs
problem
the problem states the average of four numbers is greater than or equal to one of the numbers using to find the average. we will solve this using contradiction proof.
if we take a + b + c + d / 4 = < k
we see that adding up our values, and dividing by
which is LESS than and not equal to one of the numbers being used as our average