Prove that the statement is true or false

[docnet]

Homework Statement

Consider the following statements about real numbers x and y. If true, prove it. If false, state its negation and prove that statement.

Relevant Equations

1. $\forall x>0, \exists y>0$ such that $xy<1$.
2. $\exists y>0$ such that $\forall x>0, xy<1$.

1. True Let $y=\frac{1}{x+1}$. Then $xy=\frac{x}{x+1}<1, \forall x>0$.
2. False. Let $x=\frac{1}{y}$. Then for any $x>0$, you always have $xy=\frac{y}{y}=1$

I feel like the second statement is wrong because I am choosing any $y$ , instead of choosing any $x$.

 

[fishturtle1]

It might be helpful to explicitly write the negation of 2.

 

[docnet]

It might be helpful to explicitly write the negation of 2.

Is negation an opposite statement such as "there does NOT exist a $y>0$ such that $\forall x>0, xy<1$"? So, there does not exist a $y$ that satisfies $xy<0$ for $x=\frac{1}{y}$.

 

[fishturtle1]

Is negation an opposite statement such as "there does NOT exist a y>0 such that ∀x>0,xy<1"?

Yes. But it might be even more helpful to write it in terms of quantifiers. We have:
$$\neg \exists y > 0 \forall x > 0 : xy < 1 \equiv \forall y > 0 \exists x > 0 : xy \ge 1$$

Does that make sense?

 

[sysprog]

$\neg((\exists y>0)[\forall x>0, xy<1]))\Rightarrow ((\exists z[(y=\frac 1z) \wedge(x=\frac z1)]) \Rightarrow (xy=1)))$

 

[docnet]

Yes. But it might be even more helpful to write it in terms of quantifiers. We have:
$$\neg \exists y > 0 \forall x > 0 : xy < 1 \equiv \forall y > 0 \exists x > 0 : xy \ge 1$$

Does that make sense?

Thank you, this makes sense. I have one question: does the above negation statement not carry a different meaning than the statement $\exists y > 0 \forall x > 0 : xy \ge 1$? Because, we are finding a $y$ for all $x$, as opposed of finding a $x$ for all $y$?

 

[fishturtle1]

Yes, the two statements are both false; but would require different proofs. (You might want to try to prove the one statement you wrote and compare).

as opposed of finding a x for all y?

I assume you are referring to the statement $\forall y > 0 \exists x > 0 : xy \ge 1$. For each $y$, we are finding an $x$ such that $xy \ge 1$. (I am not sure if you have seen it, but this difference comes up in continuity vs. uniform continuity).

Anyways, do you see how to finish off 2) now? (The OP is pretty close).

 

[docnet]

Yes, the two statements are both false; but would require different proofs. (You might want to try to prove the one statement you wrote and compare).


I assume you are referring to the statement $\forall y > 0 \exists x > 0 : xy \ge 1$. For each $y$, we are finding an $x$ such that $xy \ge 1$. (I am not sure if you have seen it, but this difference comes up in continuity vs. uniform continuity).

Anyways, do you see how to finish off 2) now?

Okay, that makes sense. Thank you!

2) $\exists y > 0 \forall x > 0 : xy < 1$
To prove the statement is false, we negate the statement and prove that it is true.
$\neg \exists y > 0 \forall x > 0 : xy < 1 \equiv \forall y > 0 \exists x > 0 : xy \ge 1$
Let $x=\frac{1}{y}$. Then $xy=1$. This proves the negation statement true. So its negation is false, and hence the original statement is false.

 

[fishturtle1]

edit: looks good to me.

 

[PeroK]

Thank you, this makes sense. I have one question: does the above negation statement not carry a different meaning than the statement $\exists y > 0 \forall x > 0 : xy \ge 1$? Because, we are finding a $y$ for all $x$, as opposed of finding a $x$ for all $y$?

The negation typically swaps everything in the proposition. To take an example:

The negation of $\exists x \in S: x \ \text{is odd}$ is not simply $\exists x \in S: x \ \text{is even}$. The negation is what holds iff the original statement is false. If the first statement is false (and $S$ is non-empty), then the second statement is true. But, if the second statement is true, then the first statement is not necessarily false. As there are sets with both even and odd numbers.

Instead you have to go through every member of $S$ and show that it is not odd (i.e. even). That means the negation is: $\forall x \in S: x \ \text{is even}$.

 

[docnet]

The negation typically swaps everything in the proposition. To take an example:

The negation of $\exists x \in S: x \ \text{is odd}$ is not simply $\exists x \in S: x \ \text{is even}$. The negation is what holds iff the original statement is false. If the first statement is false (and $S$ is non-empty), then the second statement is true. But, if the second statement is true, then the first statement is not necessarily false. As there are sets with both even and odd numbers.

Instead you have to go through every member of $S$ and show that it is not odd (i.e. even). That means the negation is: $\forall x \in S: x \ \text{is even}$.

Thank you. This make sense.

 

[PeroK]

The negation typically swaps everything in the proposition.

Another example that you may recognise!
$$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < |x - a| < \delta \ \text{and} \ |f(x) - f(a)| \ge \epsilon$$

 

[docnet]

Another example that you may recognise!
$$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < |x - a| < \delta \ \text{and} \ |f(x) - f(a)| \ge \epsilon$$

Is that the negation of the definition of epsilon-delta continuity?

 

[PeroK]

Is that the negation of the definition of epsilon-delta continuity?

To be precise, it says that $f$ is not continuous at the point $a$.

 

[docnet]

To be precise, it says that $f$ is not continuous at the point $a$.

Okay. To be honest, that feels like a lot of quantifiers in one statement. I am wondering, if we swapped $\exists \epsilon>0$ and $\exists x$ would the meaning of the statement change?

 

[PeroK]

Okay. To be honest, that feels like a lot of quantifiers in one statement. I am wondering, if we swapped $\exists \epsilon>0$ and $\exists x$ would the meaning of the statement change?

Yes, it would. Technically $x$ depends on $\delta$.

 

[docnet]

Yes, it would. Technically $x$ depends on $\delta$.

Okay, I understand. Thank you.

 

[PeroK]

Okay, I understand. Thank you.

To pursue this a little. Although the usual definition of continuity uses the above epsilon-delta statement, there is an equivalent alternative:

$f$ is continuous at $a$ iff for every sequence $x_n$ (in the domain of $f$ excluding $a$) that converges to $a$, the sequence $f(x_n)$ converges to $f(a)$.

It's a simple exercise the prove the equivalence of that and the usual definition.

That equivalent definition can be very useful, especially to prove that a function is not continuous. This is because the negation is simply:

$f$ is not continuous at $a$ iff there exists a sequence $x_n$ (in the domain of $f$ excluding $a$) that converges to $a$, where the sequence $f(x_n)$ does not converge to $f(a)$.

 

[sysprog]

$\neg((\exists y>0)[\forall x>0, xy<1]))\Rightarrow ((\exists z[(y=\frac 1z) \wedge(x=\frac z1)]) \Rightarrow (xy=1)))$

If it is not the case that there exists a $y$ greater than zero, such that for all $x$ greater than zero, $x$ times $y$ is less than 1, then if there exists a $z$ such that $y$ is equal to the quantity '$1$ divided by $z$', and $x$ is equal to the quantity '$z$ divided by $1$', then $x$ times $y$ is equal to $1$.