Contrapositive: Basic but Makes No Sense | xy=0, x!=0 & y!=0

[Solarfall]

This is supposedly basic but it makes no sense to me. The other topic was very old so I decided to just start a new one.

Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy!=0 then x!=0 OR y!=0.

But that is exactly false, because NEITHER x can be zero nor can y be zero else xy=0. And that is saying that "x!=0 AND y!=0", this is the only way "xy!=0".

Would someone care to justify this?

EDIT Just changing these strikes because they are very confusing. My original confusing text should still be quoted in the posts below.

 

[Dschumanji]

This is supposedly basic but it makes no sense to me. The other topic was very old so I decided to just start a new one.

Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy[STRIKE]=[/STRIKE]0 then x[STRIKE]=[/STRIKE]0 OR y[STRIKE]=[/STRIKE]0.

But that is exactly false, because NEITHER x[STRIKE]=[/STRIKE]0 nor y[STRIKE]=[/STRIKE]0 else xy=0.

Would someone care to justify this?

Let's say you have the proposition P -> Q. The contrapositive is ~Q -> ~P. In your case, P is "x=0 and y=0" and Q is "xy=0." This means that ~P becomes "x is not equal to zero or y is not equal to zero" and ~Q is "xy is not equal to zero." Therefore, the contrapositive becomes:

If xy is not equal to zero, then x is not equal to zero or y is not equal to zero. This is a true statement.

 

[BloodyFrozen]

If xy is not equal to zero, then x is not equal to zero or y is not equal to zero. This is a true statement.

Would it not be and?

 

[Dschumanji]

Would it not be and?

No.

P is of the form A and B. According to DeMorgan's Law, ~P is of the form ~A or ~B.

 

[Solarfall]

Quote the theory all you want, that doesn't make it true.

If "x!=0 or y!=0" implies one of them IS equal to zero, then you agree that xy=0? I mean that is simple.

So the question is, does or doesn't "x!=0 or y!=0" imply that one of the values is in actual fact equal to zero.

My insight regarding this is that in order for the statement to always be true, BOTH values MUST be nonzero, otherwise there IS a case in which the statement will be false.

Here is an example. If we use the conditional "x!=0 or y!=0" and substitute 5 for x and 0 for y, then "x!=0 or y!=0" will evaluate to TRUE. But x times y in that case will be 0, and so "xy!=0" will evaluate to FALSE. Which would make the statement read: If FALSE then TRUE.

Which means there is an instance, or instances, when one of the variables are equal to zero, that the Contrapositive statement is not TRUE, and is in fact fallacious.

EDIT Removed strikes

 

[gb7nash]

It looks like there's some confusion here between the converse and contrapositive.

For a statement P -> Q, the converse is Q -> P and the contrapositive is ~Q -> ~P.

Assuming that P: x = 0 or y = 0, Q: xy = 0, the converse would be:

If xy = 0, then x = 0 or y = 0. So assuming that xy = 0, at least one of these numbers must be 0. In this case, the converse is true, but like Solarfall has stated, the converse is not always true.

However, the contrapositive is always equivalent to the original statement. The contrapositive is:

If xy != 0 then x != 0 and y != 0. So assuming that xy != 0, it is a true statement that x and y are nonzero.

Edit: fixed typo

 

[awkward]

This is supposedly basic but it makes no sense to me. The other topic was very old so I decided to just start a new one.

Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy[STRIKE]=[/STRIKE]0 then x[STRIKE]=[/STRIKE]0 OR y[STRIKE]=[/STRIKE]0.

But that is exactly false, because NEITHER x[STRIKE]=[/STRIKE]0 nor y[STRIKE]=[/STRIKE]0 else xy=0.

Would someone care to justify this?

The logic is impeccable, but confusion arises because the original statement, although true, is not the strongest that can be made.

It would be better to say

If x=0 OR y=0 then xy=0.

The contrapositive of this statement is

If xy is not zero then x is not zero and y is not zero.

 

[Solarfall]

It looks like there's some confusion here between the converse and contrapositive.

For a statement P -> Q, the converse is Q -> P and the contrapositive is ~Q -> ~P.

Assuming that P: x = 0 or y = 0, Q: xy = 0, the converse would be:

If xy = 0, then x = 0 or y = 0. So assuming that xy = 0, at least one of these numbers must be 0. In this case, the converse is true, but like Solarfall has stated, the converse is not always true.

However, the contrapositive is always equivalent to the original statement. The contrapositive is:

If xy != 0 then x != 0 and y != 0. So assuming that xy != 0, it is a true statement that x and y are nonzero.

Edit: fixed typo

Forget about the converse, you are only confusing yourself.

@awkward Okay at least you realize you're not formulating exactly the same conditional statement as I am. And although I see exactly what you are saying, the statement that you mention is in fact a completely different statement, with a completely different contrapositive. In my module it is the following example, and I grasp it quite clearly and have no issues with it.

The point is simple. With this example.
If x=0 and y=0 then xy=0. This statement is ALWAYS true.
If xy!=0 then x!=0 or y!=0. This statement is NOT ALWAYS true.

QED the contrapositive of a statement is not always true.

If there is one hole, the whole thing collapses.

 

[gb7nash]

Forget about the converse, you are only confusing yourself.

Oh?

Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy[STRIKE]=[/STRIKE]0 then x[STRIKE]=[/STRIKE]0 OR y[STRIKE]=[/STRIKE]0.

This is wrong. This is the converse, not the contrapositive.

----

On a side note, if this is how you're going to respond to me trying to clear up the difference between the converse and contrapositive, find help somewhere else. Nobody's going to help you if you act like a 5 year old and try to prove your own theory that P -> Q is not equivalent to ~Q -> ~P.

 

[Char. Limit]

The statement should be:

If x=0 OR y=0 then xy=0.

This has the contrapositive:

If xy=!0 then x=!0 AND y=!0.

The converse of this would be:

If xy=0 then x=0 OR y=0.

Which is also true. All three of those statements are always true (in the real numbers).

 

[Solarfall]

Oh?



This is wrong. This is the converse, not the contrapositive.

----

On a side note, if this is how you're going to respond to me trying to clear up the difference between the converse and contrapositive, find help somewhere else. Nobody's going to help you if you act like a 5 year old and try to prove your own theory that P -> Q is not equivalent to ~Q -> ~P.

Okay I'm sorry I got carried away, it's mostly due to the strikes on this forum not really being of any help.

This is what the actual example and answer look like.
"""
Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy!=0 then x!=0 OR y!=0.
"""

Do you see it now? This is not true.

 

[Char. Limit]

Okay I'm sorry I got carried away, it's mostly due to the strikes on this forum not really being of any help.

This is what the actual example and answer look like.
"""
Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy!=0 then x!=0 OR y!=0.
"""

Do you see it now? This is not true.

You are correct, this is not always true. This is because if x=0 and y=2, then xy=0, or if x=pi and y=0, then xy=0. I think.

 

[Solarfall]

The statement should be:

If x=0 OR y=0 then xy=0.

This has the contrapositive:

If xy=!0 then x=!0 AND y=!0.

The converse of this would be:

If xy=0 then x=0 OR y=0.

Which is also true. All three of those statements are always true (in the real numbers).

How can you say what the statement SHOULD be? The statement you are referring to is the next in my examples. What would your answer be for the following statement, as is, you can't change it, it is a true statement, always:
"""
Given:
If x=0 and y=0 then xy=0.
"""
And according to theory its contrapositive is:
"""
If xy!=0 then x!=0 OR y!=0.
"""
and is supposedly always true by definition. I'm not making this up... and its wrong I tell ya. I'm giving up on anyone seeing it my way, let the fruit and vegetable hurling commence...

 

[Hurkyl]

Given:
If x=0 and y=0 then xy=0.

They say the contrapositive(which they say is always true) is:
If xy!=0 then x!=0 OR y!=0.

But that is exactly false, because NEITHER x can be zero nor can y be zero else xy=0. And that is saying that "x!=0 AND y!=0", this is the only way "xy!=0".

What is false about it?

If you know that "x!=0" is true, then it's obvious that "x!=0 or y!=0" is true.

Maybe your problem is that you are reading "or" as exclusive -- that it asserts exactly one of the two statements can be true. But that's simply wrong; in logic, "or" is inclusive. Among the cases where "P or Q" is true is the case where both P and Q are true.



All this talk about what the statement "should be" is sidetracking the discussion. All four of the following statements are true about real number arithmetic:

• If x=0 and y=0, then xy=0
• If x=0 or y=0, then xy=0
• If xy!=0, then x!=0 or y!=0
• If xy!=0, then x!=0 and y!=0

Also, by the way, "If false then true" is true.


(p.s. sorry if I've overlooked any comments. I'm rather tired and my eyes are glossing over as I try to read)

 

[Solarfall]

You are correct, this is not always true. This is because if x=0 and y=2, then xy=0, or if x=pi and y=0, then xy=0. I think.

Dude, I think I love you. No seriously, I really needed that. And I think I'm done with maths. What is more abstract then maths? Sigh, I KNEW I should have just gone and studied philosophy...

 

[Hurkyl]

Dude, I think I love you. No seriously, I really needed that. And I think I'm done with maths. What is more abstract then maths? Sigh, I KNEW I should have just gone and studied philosophy...

Char. Limit is wrong... (If I read him correctly)

 

[Char. Limit]

Oh really? So this statement:

If xy=!0, then x=!0 or y=!0

for all x and y?

Wait... that might be the case. Let me just make a truth table real quick...

 

[Solarfall]

What is false about it?

If you know that "xy!=0" is true, then it's obvious that "x!=0 or y!=0" is true.

'or' seems to imply only one of these two statements("x!=0 or y!=0") need to be true, and that one of the variables can in fact be equal to zero. In which case "xy=0". But, if

Also, by the way, "If FALSE then TRUE" evaluates to TRUE.

then I concede completely...

Thanks a lot Hurkyl, someone with a deeper understanding is what I needed here, I'm very stubborn :D

 

[Dschumanji]

A simple truth table shows that the original statement and its contrapositive are indeed equal.

 

[Solarfall]

A simple truth table shows that the original statement and its contrapositive are indeed equal.

I hate distance learning maths.

Sorry but I really didn't want to accept the textbook answer as it really didn't answer the questions I was after. But I got what I wanted. Thanks!

 

[Solarfall]

Oh really? So this statement:

If xy=!0, then x=!0 or y=!0

for all x and y?

Wait... that might be the case. Let me just make a truth table real quick...

In this case the converse of the contrapositive of the statement was in fact false, and was how I analysed it! My bad for confusing you!

 

[Solarfall]

Oh?



This is wrong. This is the converse, not the contrapositive.

----

On a side note, if this is how you're going to respond to me trying to clear up the difference between the converse and contrapositive, find help somewhere else. Nobody's going to help you if you act like a 5 year old and try to prove your own theory that P -> Q is not equivalent to ~Q -> ~P.

I was taking the converse of the contrapositive, not the converse of the original statement. Thank you for your input!

 

[Dschumanji]

The formal definitions of implication and disjunction can be confusing at times if you are just being introduced to them. I totally understand the frustration.

 

[gb7nash]

If xy!=0 then x!=0 OR y!=0.
"""

Do you see it now? This is not true.

I came back a little late, but I see what the problem is here.

"If xy!=0 then x!=0 OR y!=0". This is a true statement. "Or" in mathematic statements (unless otherwise stated) is inclusive, so there are one of three possibilites:

1) x != 0
2) y != 0
3) x != 0 and y != 0

In this case, the only way this statement is true is if 3) is satisfied, so we're fine and "If xy!=0 then x!=0 OR y!=0" is true.

This is not to be confused with exclusive or, or xor. I haven't seen any math statements that use xor.

 

[Hurkyl]

'or' seems to imply only one of these two statements("x!=0 or y!=0") need to be true, and that one of the variables can in fact be equal to zero.

This, I think, is one of the bigger departures of formal language from natural language.

Given its normal purpose, natural language tries to be efficient -- it would be wasteful to talk about things that are irrelevant. Thus an English statement like "If X, then Y" contains the extra mild implication that X is actually true, at least sometimes. A clause "X or Y" mildly implies that X happens sometimes and Y happens sometimes. And there are other natural language nuances to words like "or", "and", "if ... then", and so forth.

Alas, while this is fine for hunting animals in the jungle, it's not so good for doing mathematics.


That said, if a mathematician proved a theorem "If X, then Y or Z", they would typically be inspired to do extra work, looking for examples of:

• X true, Y true, Z false
• X true, Y false, Z true
• X true, Y true, Z true

and use them to demonstrate that there are no obvious improvements to the statement of the theorem. (Or, upon failure to find an example, seek to find a proof of the suggested improvement of the theorem) He would also seek to refine X to split it into two cases on which he could say "If X1 then Y" and "If X2 then Z".

(Of course, seeking to do these things doesn't always mean that they'll be successful, but anyways)

 

[Solarfall]

This, I think, is one of the bigger departures of formal language from natural language.

Given its normal purpose, natural language tries to be efficient -- it would be wasteful to talk about things that are irrelevant. Thus an English statement like "If X, then Y" contains the extra mild implication that X is actually true, at least sometimes. A clause "X or Y" mildly implies that X happens sometimes and Y happens sometimes. And there are other natural language nuances to words like "or", "and", "if ... then", and so forth.

Alas, while this is fine for hunting animals in the jungle, it's not so good for doing mathematics.


That said, if a mathematician proved a theorem "If X, then Y or Z", they would typically be inspired to do extra work, looking for examples of:

• X true, Y true, Z false
• X true, Y false, Z true
• X true, Y true, Z true

and use them to demonstrate that there are no obvious improvements to the statement of the theorem. (Or, upon failure to find an example, seek to find a proof of the suggested improvement of the theorem) He would also seek to refine X to split it into two cases on which he could say "If X1 then Y" and "If X2 then Z".

(Of course, seeking to do these things doesn't always mean that they'll be successful, but anyways)

I'm not sure if you are calling me a Neanderthal or a mathematician, or both for that matter.

I think language and maths are more closely related than you are pointing to here though. The language itself doesn't imply anything, it is the user of the language that does this.

I'm very glad to stand corrected though!

 

[Dschumanji]

I'm not sure if you are calling me a Neanderthal or a mathematician, or both for that matter.

I think language and maths are more closely related than you are pointing to here though. The language itself doesn't imply anything, it is the user of the language that does this.

I'm very glad to stand corrected though!

Hurkyl isn't name calling. He is just saying that the evolution of natural languages are not precise enough at times when it comes to mathematical reasoning.

He never said that mathematics and language are loosely related. In fact I think his post says the opposite. He shows how precise language is necessary for mathematics. This is why formal languages are used in mathematical reasoning. Natural languages and formal languages can appear very similar on the surface and this is where most of the confusion about logical and mathematical statements comes from: people tend to interpret what they see with the natural language they are comfortable using.

Natural languages do imply certain things. Hurkyl gave a great example with implications.

 

[Solarfall]

Hurkyl isn't name calling. He is just saying that the evolution of natural languages are not precise enough at times when it comes to mathematical reasoning.

He never said that mathematics and language are loosely related. In fact I think his post says the opposite. He shows how precise language is necessary for mathematics. This is why formal languages are used in mathematical reasoning. Natural languages and formal languages can appear very similar on the surface and this is where most of the confusion about logical and mathematical statements comes from: people tend to interpret what they see with the natural language they are comfortable using.

Natural languages do imply certain things. Hurkyl gave a great example with implications.

I wasn't offended, I was kidding. Thanks.

 

[paulfr]

• If x=0 and y=0, then xy=0
• If x=0 or y=0, then xy=0
• If xy!=0, then x!=0 or y!=0
• If xy!=0, then x!=0 and y!=0

Hurky

Isn't the 3rd statement here false ?
If xy not = 0 then x and y are not = 0
So x not = 0 OR y not = 0 is not sufficient for xy not = 0
Both x and y must not be zero for xy not= 0

 

[Hurkyl]

Hurky

Isn't the 3rd statement here false ?
If xy not = 0 then x and y are not = 0
So x not = 0 OR y not = 0 is not sufficient for xy not = 0
Both x and y must not be zero for xy not= 0

You're confusing it with its converse. The following statement is indeed not a theorem:

If $x \neq 0$ or $y \neq 0$ then $xy \neq 0$​

and the same can be said for its contrapositive

If $xy = 0$ then $x = 0$ and $y = 0$​