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The one rearrangement of an if-then statement that says the same thing, the two that do not, and the two argument forms that go wrong by using them.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write down the converse, inverse and contrapositive of any implication, say which of them is equivalent to the original and show with a truth table why, translate the phrases that hide an arrow — only if, unless, necessary, sufficient — into symbols, and name the two argument forms that are valid and the two that are not. You will also be able to say what single case breaks affirming the consequent, which is the case worth keeping in your head for the rest of the course.
From lesson 1 you have the implication's table: false in exactly one row, the row where the hypothesis holds and the conclusion fails. Everything here is read off that single fact, so if the four rows are not solid, fill them in again before going on.
Given $P \Rightarrow Q$: the converse is $Q \Rightarrow P$, the inverse is $\neg P \Rightarrow \neg Q$, and the contrapositive is $\neg Q \Rightarrow \neg P$. A condition is sufficient for a claim when it implies it, and necessary when the claim implies it.
Take $P \Rightarrow Q$ and rearrange it in the three available ways. Two of the results are the original claim in other words and two are a different claim, and the table settles which.
| Formula | Name | False in the row | Same claim? |
|---|---|---|---|
| $P \Rightarrow Q$ | the statement | $P$ true, $Q$ false | — |
| $Q \Rightarrow P$ | converse | $Q$ true, $P$ false | no |
| $\neg P \Rightarrow \neg Q$ | inverse | $P$ false, $Q$ true | no |
| $\neg Q \Rightarrow \neg P$ | contrapositive | $P$ true, $Q$ false | yes |
The contrapositive is ruled out by exactly the row the original is ruled out by, so the two are the same statement with the letters moved. That is not a curiosity: it is a licence. To prove $P \Rightarrow Q$ you may instead prove $\neg Q \Rightarrow \neg P$, and unit 2 spends a whole lesson on when that is the easier road.
The converse and the inverse are each other's contrapositive, so they are the same claim as one another and neither follows from the original. 'Differentiable implies continuous' is a theorem; its converse is false, and $|x|$ at zero is why.
Necessary and sufficient are the same arrows under other names. $Q$ is necessary for $P$ when $P \Rightarrow Q$: the claim cannot hold without it. $Q$ is sufficient for $P$ when $Q \Rightarrow P$: it is enough on its own. Both at once is $P \Leftrightarrow Q$, which is what a definition asserts.
Another way: steps
To decide what a rearrangement is:
Another way: example
'The lights stay on unless the fuse blows' has no arrow visible. Unless is if not, so it is $\neg Q \Rightarrow P$ with $P$ the lights and $Q$ the fuse — and its contrapositive, $\neg P \Rightarrow Q$, is the useful form: the lights went out, so the fuse blew.
Two premises and a conclusion, over and over:
| Premises | Conclusion | Name | Verdict |
|---|---|---|---|
| $P \Rightarrow Q$, $P$ | $Q$ | modus ponens | valid |
| $P \Rightarrow Q$, $\neg Q$ | $\neg P$ | modus tollens | valid |
| $P \Rightarrow Q$, $Q$ | $P$ | affirming the consequent | invalid |
| $P \Rightarrow Q$, $\neg P$ | $\neg Q$ | denying the antecedent | invalid |
The two valid moves use the statement and its contrapositive. The two invalid ones use the converse and the inverse, and both are broken by the same single case: $P$ false, $Q$ true. It is worth checking that case by hand once — the premises come out true, the conclusion false, and that is the whole of what invalid means.
Proving the converse. A proof of the converse looks exactly like a proof of the statement until you check which way the assumption ran. Before writing a line, say out loud what you are assuming and what you are trying to reach.
*Reading only if as if. 'The series converges only if the terms tend to zero' says convergence requires* the terms to vanish. It does not say vanishing terms give convergence, and the harmonic series is the standing counterexample.
Treating a necessary condition as a test. Being even is necessary for being divisible by six and settles nothing on its own; $4$ passes the test and fails the claim.
It is a different statement, and it can be false while the original is true or true while the original is false. The pair 'differentiable $\Rightarrow$ continuous' and its converse is the standard example in this direction; '$x > 2 \Rightarrow x > 1$' and its converse is one in the other. Nothing about the original licenses any opinion about the converse, which is why a theorem and its converse are always proved separately and often only one of them is true.
Claim: if $n^2$ is even then $n$ is even. Assuming '$n^2$ is even' gives you almost nothing to write down.
Notice that the hypothesis is unhelpful.
The contrapositive is: if $n$ is odd then $n^2$ is odd. Now the hypothesis is concrete — $n = 2k+1$.
Swap and negate both halves.
$n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, odd. Because the contrapositive is the same claim, the original is proved.
One line, because the assumption was usable.
'If it rained, the ground is wet. The ground is wet. So it rained.' Write it: $P \Rightarrow Q$, $Q$, therefore $P$.
Symbols before judgement.
That is affirming the consequent — the converse in disguise. Take $P$ false and $Q$ true: both premises hold and the conclusion fails.
One row breaks it.
In words: somebody washed the car. The argument is invalid even on the days it happens to rain, because validity is about the form and not about the weather.
Invalid is about form, not about truth.
Sufficient means enough on its own, so the condition is the hypothesis: $Q \Rightarrow P$ with $Q$ the determinant and $P$ invertibility.
Sufficient marks the hypothesis.
Its contrapositive is $\neg P \Rightarrow \neg Q$: a matrix that is not invertible has determinant zero. Same claim, and it is the form you would actually use.
Swap and negate.
Its converse, $P \Rightarrow Q$, is a separate claim — here also true, which is why the two are usually stated together as an if and only if. That the converse happens to be true is something you have to prove, not something the first sentence gave you.
Take $P \Rightarrow Q$. Complete the column for $P \Rightarrow Q$, the statement itself. The rows run: both true, $P$ true and $Q$ false, $Q$ true and $P$ false, neither.
| P | Q | P -> Q |
|---|---|---|
Premises: P | Q and ~P. Conclusion: Q. Is the argument valid? If not, give a case that breaks it.
P | Q
~P
∴ Q
valid invalid — countermodel:
Take '**if** a number is divisible by $4$ **then** it is even'. What is 'if it is not even then it is not divisible by $4$'?
Let P stand for “the lights stay on” and Q for “the fuse blows”. Write this in symbols: “the lights stay on unless the fuse blows.”
Answer:
For an integer $n$, sort each condition by what it gives you about '$n$ is divisible by $6$'.
| Necessary, but not sufficient | Sufficient, but not necessary | Necessary and sufficient | Neither | |
|---|---|---|---|---|
| $n$ is even | ||||
| $n$ is divisible by $12$ | ||||
| $n$ is divisible by $2$ and by $3$ | ||||
| $n$ is positive |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Take $P \Rightarrow Q$. Complete the column for $P \Rightarrow Q$, the statement itself. The rows run: both true, $P$ true and $Q$ false, $Q$ true and $P$ false, neither.
| P | Q | P -> Q |
|---|---|---|
You can rearrange an implication four ways and say which rearrangements are the same claim. Say in your own words why the contrapositive is equivalent and the converse is not, and name the one case that breaks both fallacies. Next: when two formulas count as the same statement, and the laws that let you rewrite one as another.
10. Your turn: what does 'a non-zero determinant is sufficient for invertibility' assert?, step 3