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For the following statement A implies B, write the statement NOT B implies NOT A.

Let $a$, $b$, and $c$ be real numbers. If $a > 0$, then there does not exist a real number $M$ such that, for every real number $x$, $ax^2 + bx + c \le M$.


Textbook Solution

If there exists a real number $M$ such that, for every real number $x$, $ax^2 + bx + c \le M$, then $a \le 0$.


My Solution

If there exists real numbers $M$ and $x$ such that $ax^2 + bx + c > M$, then $a \le 0$.


As I understand it, NOT [for every] and NOT [$ax^2 + bx + c \le M$] should be equivalent to [there exists] and [$ax^2 + bc + c > M$]; indeed, this has been the case with all previous similar questions.

I'm wondering what I've misunderstood here, or is my solution correct?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ I think you're incorrectly applying the NOT to all the terms in the statement, when it should just apply to the initial “there does not exist”. ¬(∄ M: (∀ x ∈ ℝ: (ax² + bx + c ≤ M))) ↔ ∃M: (∀ x ∈ ℝ: (ax² + bx + c ≤ M)). Think of it like multiplication: if you negate i(j + k), you get -i(j + k), not -i(-(j + k)). $\endgroup$ – David Moles Feb 7 '17 at 20:28
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The logical form of the implication you're given is $a > 0 \implies\neg \exists M \alpha$, where $\alpha$ just stands for the second inequality that you wrote in your question. The contrapositive is $\exists M \alpha \implies a \leq 0$.

The main thing to recognize is that the consequent of the given conditional is a negated existential sentence: "There does not exist ..." In the contrapositive, this sentence becomes an ordinary existential sentence: "There exists..."


Added: You seem to be confusing quantifier equivalences and contrapositives. It's true that $\neg \exists \alpha \equiv \forall \neg \alpha$ and dually $\neg \forall \alpha \equiv \exists \neg \alpha$. But we needn't use those equivalences here.

We use the contrapositive equivalence $A \implies B \equiv \neg B \implies \neg A.$

Now, again, we are given $a > 0 \implies \neg \exists M \alpha$. By the contrapositive equivalence this becomes $\neg \neg \exists M \alpha \implies \neg(a > 0)$. We can remove the double negation, and use $\neg(a>0) \equiv a \leq 0$ to get what I wrote above (which agrees with the textbook answer).

The thing to notice is that, in the contrapositive sentence, there is no longer any negation to "push in" over a quantifier, so the quantifier equivalences that I wrote above do not apply.

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  • $\begingroup$ What I don't understand is why the second quantifier (for all) is not changed to "there is"? My textbook indicates that I should be changing it: "If the word NOT appears to the left of a quantifier, then move the word NOT to the right of the quantifier and place the NOT just before the something that happens. As you do so, change the quantifier to its opposite—“for all” becomes “there is” and “there is” becomes “for all.” Repeat this step for nested quantifiers so long as the word NOT appears to the left of a quantifier." $\endgroup$ – The Pointer Feb 7 '17 at 18:48
  • $\begingroup$ Can you clarify what the first quantifier is? $\endgroup$ – grndl Feb 7 '17 at 18:50
  • $\begingroup$ The first quantifier would be "there does not exist"; the second would be "for every". According to the textbook, I should be converting these both to "there exists" and then changing the inequality? $\endgroup$ – The Pointer Feb 7 '17 at 18:51
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    $\begingroup$ I've expanded my answer. Hope that helps. $\endgroup$ – grndl Feb 7 '17 at 18:55
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    $\begingroup$ Not quite. Simply $A \implies \neg B \equiv B \implies \neg A$. This is just another instance of the contrapositive equivalence. $\endgroup$ – grndl Feb 7 '17 at 19:10
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To focus on the part that's confusing you, I suggest thinking about the latter part as a statement about the number $M$: specifically, let's declare $P_{a,b,c}(M)$ to be the statement

"$ax^2+bx+c\leq M$ for every real number $x$."

So the original statement is: "If $a>0$, then there does not exist any $M$ for which $P_{a,b,c}(M)$ is true."

Then the contrapositive statement is: "If there does exist an $M$ for which $P_{a,b,c}(M)$ is true, then $a\leq 0$".

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  • $\begingroup$ Shouldn't it be $P_{a,b,c}(M, x)$? Also, in setting up this declaration, I think it would be fair to also mention the proper domains. $\endgroup$ – ThisIsNotAnId Feb 7 '17 at 19:10
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    $\begingroup$ @ThisIsNotAnId: No - in the way I am phrasing things, it is not a statement about specific numbers $M$ and $x$, it is a statement about the number $M$ which involves all $x$. I also felt it would unnecessarily complicate things for the OP to deal with the domains here. $\endgroup$ – Zev Chonoles Feb 7 '17 at 19:22
  • $\begingroup$ Yes I was thinking maybe domains will clear this up, but you have a good point. In that light, my answer just expands upon yours. Although I thought that setting it up the way I did might be helpful, albeit tedious. $\endgroup$ – ThisIsNotAnId Feb 7 '17 at 19:23
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Let $a, b, \text{and } c$ be real numbers. If $a>0$, then there does not exist a real number $M$ such that, for every real number $x$, $ax^2+bx+c≤M$.

Let's break this down. The first part just tells us that we are working with real numbers. This will be irrelevant to the discussion as those are the only kind of numbers we are working with, so any logical manipulation will have to be with real numbers.

The "if" part of the statement: $A = a>0$. So right away we know the negation of this is $a \leq 0$.

Now, the "then" part. The claim is that for every real number $x$, there does not exist a real number $M$ such that... In other words, there is no single $M$ such that ... for every real number $x$. Or even more simply and ambiguously, there is no "one-size-fits-all" $M$ for all the real numbers. And the condition is that $ax^2+bx+c≤M$. Just to summarize the "then" part has this format: "No single $M$ for all the $x$'s such that this happens: $ax^2+bx+c≤M$." Summarizing it this way makes it easy to come up with the negation. I just have to come up with a "generic" counter-example.

The negation to the "then" part is: There is a single $M$ for all the $x$'s such that this happens: $ax^2+bx+c≤M$. Let this statement be called $\neg B = C$.

Now you just put it together: (If $\neg B, \text{then } \neg A) \iff (C \implies \neg A)$.

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Let $a, b,$ and $c$ be real numbers. If $a>0$, then there does not exist a real number $M$ such that, for every real number $x$, $ax^2 +bx+c≤M$ .

The statement is: $$a>0 ~\to~\neg\color{silver}{\Big(}\exists M{\in}\Bbb R~\color{silver}{\big(}\forall x{\in}\Bbb R~(ax^2+bx+c\leqslant M)\color{silver}{\big)\Big)}$$

A contraposition of a conditional switches position of its antecedent and consequent, and negates each of them.   This negation is applied to the whole phrase, not individually to nested components.

$$\neg\neg\color{silver}{\Big(}\exists M{\in}\Bbb R~\color{silver}{\big(}\forall x{\in}\Bbb R~(ax^2+bx+c\leqslant M)\color{silver}{\big)\Big)}~\to~\neg(a>0)$$

Although in some cases you may "move" the negation to the interior using Duality rules, and such, it is safest not to do so in your head unless you are justly confident about your mental legerdemain.   Take the thyme to do it write and knot make any mistakes.

In this case we can instead immediately use Double Negation elimination: $$\color{silver}{\Big(}\exists M{\in}\Bbb R~\color{silver}{\big(}\forall x{\in}\Bbb R~(ax^2+bx+c\leqslant M)\color{silver}{\big)\Big)}~\to~(a\leqslant 0)$$

"If there exists a real number $M$ such that for all real values $x$ we have $ax^2+bx+c\leqslant M$, then $a$ is non-positive."

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