# Prove that the statement is true or false

• docnet
In summary, the negation of the statement ##\exists y > 0 \forall x > 0 : xy < 1## is false because it always has the result ##1##.
docnet
Gold Member
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##.

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

benorin and docnet
fishturtle1 said:
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}##.

docnet said:
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?

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

Last edited:
docnet
fishturtle1 said:
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##?

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).

docnet said:
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
fishturtle1 said:
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.

PeroK and fishturtle1
edit: looks good to me.

docnet
docnet said:
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
PeroK said:
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 said:
The negation typically swaps everything in the proposition.
Another example that you may recognise!
$$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < |x - a| < \delta \ \text{and} \ |f(x) - f(a)| \ge \epsilon$$

PeroK said:
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?

docnet said:
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##.

PeroK said:
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?

docnet said:
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##.

PeroK said:
Yes, it would. Technically ##x## depends on ##\delta##.
Okay, I understand. Thank you.

docnet said:
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)##.

docnet
sysprog said:
##\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##.

docnet

## 1. How do you prove a statement to be true or false?

In order to prove a statement to be true or false, you must use evidence and logical reasoning to support your argument. This can include conducting experiments, analyzing data, and making logical deductions based on established theories and principles.

## 2. What is the difference between proving a statement to be true or false?

The main difference between proving a statement to be true or false is the approach and evidence used. To prove a statement to be true, you must provide evidence that supports its validity. To prove a statement to be false, you must provide evidence that contradicts or disproves the statement.

## 3. Can a statement be proven to be both true and false?

No, a statement cannot be both true and false. A statement is either true or false based on the evidence and logical reasoning used to support it. If there is evidence to support a statement, then it can be proven to be true. If there is evidence to contradict a statement, then it can be proven to be false.

## 4. What role does experimentation play in proving a statement?

Experimentation is a crucial part of proving a statement to be true or false. By conducting experiments, scientists can gather data and evidence to support their argument. This data can be used to make logical deductions and reach a conclusion about the truth or falsity of a statement.

## 5. Is it possible to prove a statement without any evidence?

No, it is not possible to prove a statement without any evidence. Evidence is necessary to support any argument or claim. Without evidence, a statement cannot be proven to be true or false. It is important for scientists to use reliable and valid evidence in order to make accurate conclusions about the truth or falsity of a statement.

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