Only brainlets think that if the an intitial theorem is true, then the contrapositive is always true.
Consider this: If a is an odd integer, and ab is an odd integer, then b is an odd integer.
this is easily shown to be true. But what if we take the contraposition of it?
Consider the CP: If b is not an odd integer then a is not an odd integer or ab is not an odd integer
knowing that b is not an odd integer tells us nothing about a the properties of a, so lets see what happens when we try to prove that ab is not an odd integer
let b =1/2 let a = 2 therefore ab = 1 which is an odd integer
thus the contrapositive has been shown to be false brainlets BTFO
I know this is bait, but fuck it. 1/2 is not an integer so try again
Charles Anderson
that was the premise of the contrapositive, brainlet. b is not an odd integer 1/2 is not an odd integer. the contrapositive is false, brainlet. try again.
Chase Myers
>If a is an odd integer, and ab is an odd integer, then b is an odd integer. This is wrong. let a=3 and ab=5. a=3 is odd. ab=5 is odd. b=5/3 is not an odd integer.
Evan Johnson
ding ding ding
Bentley Brooks
Just write down a truth table you fucking retard
Blake King
you said it yourself, te contrapositive is "if b is not an odd integer then a is not an odd integer or ab is not an odd integer"
in your example a is not an odd integer, so you're not showing that the contrapositive is false.
Adrian Gonzalez
Let a and b integers, then (a odd) and (ab odd) => (b odd). Cp : (b even) => (a even) or (ab even). If b is even, then ab is even, so it's true. I took a and b integers because that's easier, but it doesn't change anything.
Robert Allen
>knowing that b is not an odd integer tells us nothing about a the properties of a But it does tell you something about the relation between the properties of a, and the properties of ab.
>so lets see what happens when we try to prove that ab is not an odd integer You can't, for that is not what the contrapositive claims.
The contrapositive claims that a is not an odd integer, or ab is not an odd integer. You can only prove the disjunction, not any individual leg of it.
>let b =1/2 >let a = 2 >therefore ab = 1 which is an odd integer Here, a is not an odd integer, so "a is not an odd integer or ab is not an odd integer" holds, as promised.
>thus the contrapositive has been shown to be false It has not.
Gavin Green
Assume "A => B" and we know that "!B" holds
Then there are two possibilities for A: A and !A Since we know A implies B, if A, then also B, which is a contradiction to the fact that we already know !B Therefore !A
Gabriel Mitchell
ding ding the initial theorem was false to begin with
Matthew Jackson
>Only brainlets think that if the an intitial theorem is true, then the contrapositive is always true. Ok then. Suppose the second theorem does not follow from the first.
If A => B is true, then B is true when A is true (by definition).
Suppose that B is false, and A is true. If A is true, then B is true. But it is also false. Therefore A and the the contrapositive of B cannot be simultaneously true. Therefore if we maintain that -B is true, -A must also be true. Thus it follows that -B => -A if A => B
Austin Cook
Since the two statements are equivalent, we know !b is false.
!b implies !a or !ab is only false if !b is true and !a or !ab is false.
Since !b is false, the contrapositive is true for all a,b such that !a and !ab
>he can't tell the difference between semantical tautology and formal derivation
Eli Johnson
>She's using her very own fantasy rules >hand wavy maximization
Owen Flores
Mate the first logical statement only makes sense if it's integers only. Taking the contrapositive of that doesn't mean that you can use imaginary numbers. a only implies b if you are using integers. Your initial proof only works if it's integers >Hence this The logical implication requires the math be restricted to integers and therefore the contrapositive of the logic must also be restricted.
Blake Gomez
>t. True intelligence He just calls people retards for fun in his spare time without justification on a board dedicated to math.
Jace Young
>without justification because some poeple, as early as the 3rd post, already did and retards just go on.
Joshua Collins
>Consider the CP Reported
Colton Gomez
Contrapositive is: If b is not an odd integer then a is not an odd integer or ab is not an odd integer
If b = 1/2, a = 2, then b is not an odd integer, and a is also not an odd integer, so "a is not an odd integer OR ab is not an odd integer" is true. So the contrapositive is true, try again.
Jace Kelly
If you allow a and b to be non-integers, then neither the original statement nor its contrapositive is true. e.g. a = 3, b=1/3. Your counterexample does not work though. So this is probably bait.
Joseph Reed
goddamn brainlet the first statement isn't true, you can't conclude that if a is an odd integer, and if ab is an odd integer, then b is an odd integer you need a third statement >if a is an odd integer, and if b is an integer, and if ab is an odd integer, then b is an odd integer now, when you take the contrapositive, you get >if b is not an odd integer, then a is not an odd integer, or b is not an integer, or ab is not an odd integer which is true.
Jayden Lee
in fact, the first statement could be generalized further, and you would get the statment >if a is an integer, and if b is an integer, and if ab is an odd integer, then a is an odd integer, and b is an odd integer which gives us the contrapositive >if a is not an odd integer, or if b is not an odd integer, then a is not an integer, or b is not an integer, or ab is not an odd integer
