I have actually a vague question about proof by contradiction. The main goal that proofs by contradiction is to present that if part statement to be true, then we would finish up through a logical contradiction. My question would: in stimulate to achieve the logical contradiction, would certainly it be enough to just show a counter-example? the is we mean the false statement, and also then just give an example where the statement provides an illogical/absurd result. Would that be correct?  No that"s not enought (if i taken what you mean). Think about it this way:

You want to prove A => B and you desire to carry out it through contradiction. That means you suppose B is false also with the theory A. Offering a counterexample to this way giving an example of A => B, and this doesn"t prove A=>B in general.

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For example:I try to prove the every strange number is prime (which is false);By contradiction expect n is odd yet not prime. Ns can discover a counterexample to this (3 is odd and also prime) however this doesn"t offer me the thesis (in reality 9 is odd yet not prime) What you"re describing sounds much more to me favor a proof by counterexample than a proof by contradiction. Because that instance, in Mauro"s proof the the statement "every natural number is even," the didn"t should assume because that the benefits of contradiction that no every number is even. He simply pointed to the number 3.

Some so-called proofs by contradiction are proofs by contrapositive instead. Come prove \$Aimplies B\$, part writers will assume \$A\$ and also \$ eg B\$, climate derive \$ eg A\$, and also say \$A\$ and \$ eg A\$ is a contradiction. Yet what they have actually actually done is proven \$ eg B implies eg A\$, i m sorry is the contrapositive the \$Aimplies B\$.

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A genuine proof through contradiction will certainly arrive at some kind of internal contradiction. A very great example is the proof that \$sqrt2\$ is irrational. You i think it is rational, then compose it in shortest terms as \$fracab\$. Indigenous \$a^2 = 2b^2\$, you derive that both \$a\$ and \$b\$ space divisible through \$2\$, for this reason \$fracab\$ can"t be in lowest terms after all. So friend see, the contradiction doesn"t have to do with the statement in the theorem, so much as miscellaneous that would follow from that statement being false.