What is the correct way of disproving a mathematical statement? Assuming the thing you want to disprove and inferring something you know to be false This is called reductio ad absurdum, or proof by contradiction It may seem bizarre to assume a (possibly) false proposition, but what you are really doing is considering the truth value of propositions conditional on the proposition you want to disprove
Proving Disproving Product of two irrational number is irrational In fact, $\lnot q\to\lnot p$ means [Rational number -> Cannot be broken down into product of two irrational numbers] (which is a false statement, naturally, and not easier to disprove than the original statement, but this is another matter)
Proving Functions are Surjective - Mathematics Stack Exchange You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
For all integers a, b, and c, if ab | c then a | c and b | c Prove or disprove the following: For all integers a, b, and c, if ab | c then a | c and b | c I'm having trouble proving the above It seems to be obviously true in my head (but only because of a
Is this a valid proof? If $a$ and $b$ are rational, $a^b$ is rational. Technically, your proof is fine You showed that for all a and b are rational numbers, a^b is not always rational, by counterexample Unless you're going to prove the statement by contraposition, finding the contrapositive is unnecessary Just for future reference, the contrapositive of "if a and b then d", is "if ~d then ~a or ~b" (this is due to DeMorgan's Law)