Proof by contradiction

[arhzz1]

Hello!

I am having a hard time understanding the proof of condtradiction, the logic you follow to get to the correct solution.

This is my problem:

∀x from R : |x − 4| < 1 ⇒ x < 5

Now I did the step where you negate the B part of the statement;

|x - 4| ^ x>= 5

But now I am stuck, not sure what my next step is. If anyone can help would be much appreciated.

PS If this problem isn't in the right category I am very sorry I've googled proof of contradiction too see in which math category it belongs in but with no success.

 

[Steven G]

Well you can't prove this statement as it is not true.

You claim x<5. OK, so x=-10 should work. |-10 - 4| = |-15| = 15 but this is not less than 1.

If |x-4|<1 then the distance between x and 4 will be less than 1. That only works if x is in between 3 and 5. That is 3<x<5, NOT x<5

Also you talk about where you negate the B part of the statement. Can you please tell us what the B part of the statement means? Is there an A part?

 

[arhzz1]

A

Well you can't prove this statement as it is not true.

You claim x<5. OK, so x=-10 should work. |-10 - 4| = |-15| = 15 but this is not less than 1.

If |x-4|<1 then the distance between x and 4 will be less than 1. That only works if x is in between 3 and 5. That is 3<x<5, NOT x<5

Also you talk about where you negate the B part of the statement. Can you please tell us what the B part of the statement means? Is there an A part?

Okay maybe I put the question wrong, I'm actually susposed to check if the statement is true or false, so you saying I cannot prove it right, sorry for the incovinience. I've solved this another way (I cannot say it in english but "Fallunterscheidung" in german) and I also get that the statement works only for the numbers between 3 and 5. Now about the A and B part In my script there is a formula:

(A ⇒ B) ≡ (A ∧ ¬B ⇒ F)

This is what I meant by A and B part; A should be |x-4| <1 and B the part after the ⇒. So if I've gotten this correctly, the only trick to solving this with a proof of contradiction is just finding 1 example that doesnt work for my assumption? But what if the statement was more complex, not something you could just put in and calculate.Is there any "path" you take towards making the right assumption or is it just expirience?

Thank you for your answer btw. and sorry for the confusion in the question.

Cheers mate!

 

[Dr.Peterson]

Well you can't prove this statement as it is not true.

You claim x<5. OK, so x=-10 should work. |-10 - 4| = |-15| = 15 but this is not less than 1.

If |x-4|<1 then the distance between x and 4 will be less than 1. That only works if x is in between 3 and 5. That is 3<x<5, NOT x<5

No, no, no!

The claim is not that if x<5, then |x − 4| < 1 . It's that if |x − 4| < 1, then x<5.

 

[Dr.Peterson]

∀x from R : |x − 4| < 1 ⇒ x < 5

Now I did the step where you negate the B part of the statement;

|x - 4| ^ x>= 5

But now I am stuck, not sure what my next step is. If anyone can help would be much appreciated.

Did you mean to write, |x - 4| < 1 ^ x >= 5 ?

Now you need to show that these can't both be true. For example, if x >= 5, then what can you say about |x - 4|?

(By the way, I don't even look at the category, so I wouldn't worry about it; the algebra here is fairly basic, but you're probably in a course about proofs that goes beyond algebra, so Advanced Algebra or even Advanced Math could work.)

 

[arhzz1]

Did you mean to write, |x - 4| < 1 ^ x >= 5 ?

Now you need to show that these can't both be true. For example, if x >= 5, then what can you say about |x - 4|?

(By the way, I don't even look at the category, so I wouldn't worry about it; the algebra here is fairly basic, but you're probably in a course about proofs that goes beyond algebra, so Advanced Algebra or even Advanced Math could work.)

Did you mean to write, |x - 4| < 1 ^ x >= 5 ?

Yes I did mean to write that, my bad (AGAIN). Now if we say that x is greater than 5, we can input a value of lets say 7 and when we put that into |x-4| we get 7-4 < 1; 3<1 which makes no sense, thus making the statement not true?

 

[pka]

Did you mean to write, |x - 4| < 1 ^ x >= 5 ?
Now you need to show that these can't both be true. For example, if x >= 5, then what can you say about |x - 4|?
(By the way, I don't even look at the category, so I wouldn't worry about it; the algebra here is fairly basic, but you're probably in a course about proofs that goes beyond algebra, so Advanced Algebra or even Advanced Math could work.)

I completely agree with the above. I don't see how this is Basic Algebra.
You have been asked to prove \((\forall x)[P(x)\Rightarrow Q(x)]\) by contradiction.
Here \(P(x)\text{ is }|x-4|<1\) and \(Q(x)\text{ is } x<5\).
The negation of the statement, \(\neg(\forall x)[P(x)\Rightarrow Q(x)] \equiv (\exists t)[P(t)\wedge \neg Q(t)]\)
So assume (\(\exists t)[|t-4|<1\)\( \wedge t\ge 5]\) but that means both are true:
\(|t-4|<1\text{ means }\\-1<t-4<1\\3<t<5\) but we assume that \(\bf t\ge 5\). there is the contradiction.

 

[arhzz1]

I completely agree with the above. I don't see how this is Basic Algebra.
You have been asked to prove \((\forall x)[P(x)\Rightarrow Q(x)]\) by contradiction.
Here \(P(x)\text{ is }|x-4|<1\) and \(Q(x)\text{ is } x<5\).
The negation of the statement, \(\neg(\forall x)[P(x)\Rightarrow Q(x)] \equiv (\exists t)[P(t)\wedge \neg Q(t)]\)
So assume (\exists t)[|t-4|<1 \wedge 7\ge 5]\) but that means both are true:
\(|t-4|<1\text{ means }\\-1<t-4<1\\3<t<5\) but we assume that \(\bf t\ge 5\). there is the contradiction.

Im sorry but I dont understand this part; Im guessing something went wrong with the formmating?

So assume (\exists t)[|t-4|<1 \wedge 7\ge 5]\) but that means both are true:

 

[pka]

Im sorry but I dont understand this part; Im guessing something went wrong with the formmating?

So assume (\exists t)[|t-4|<1 \wedge 7\ge 5]\) but that means both are true:

I corrected it.

 

[arhzz1]

I corrected it.

I see now, thank you for the great answer (both of you) I've understood now

 

[Steven G]

No, no, no!

The claim is not that if x<5, then |x − 4| < 1 . It's that if |x − 4| < 1, then x<5.

I had my doubts about my last post. I am having a bad day. Thanks for correcting me mostly for the OP.