Discrete Mathematics proof question help

promitheus

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Aug 19, 2017
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Hi,

I'm struggling to get my head around this particular example.

Q: Find the error in the following \proof" that 10 = 20. Can you express the incorrect step in terms of an invalid argument?

Suppose that 10 = 20. Then subtracting 15 from both sides we get -5 = 5, and squaring both sides we get (-5)2 = 25 = 52, which is true. Therefore our original assumption that 10 = 20 must be true too.

My working/thought process so far:

Let p be the proposition that 10=20

Let q be the manipulations (the subtracting 15 from both sides and squaring)

These are the two premises and the conclusion is that if q, then p (q->p). In order for our conclusion to hold true, both our premises must also be true, and as our proposition p has a truth value of 0, our conclusion cannot be true. Therefore, the argument is invalid.

I'm really unsure of this because the assumption is so obviously wrong, however I'm not sure if my method of proving it is correct?

Would appreciate any feedback.

Cheers.
 

Dr.Peterson

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You can't really say that proposition q is "the manipulations". A proposition must be a specific statement, such as "(-5)^2 = 5^2". Let's use that for our q.

Also, you can't call "manipulations" a "premise". Neither your q nor my q is something you are assuming for the sake of argument. Note that the conclusion of the argument is that 10 = 20, which is your p, not q -> p or p -> q.

The manipulations show that p -> q, right? So the reasoning is that if p -> q, and q, then p.

Is that a valid argument? Or is it one of the classic fallacies you have learned about?

And how does the reasoning here compare to, say, a proper proof by contradiction?

(For more help, it would be good if you could tell us what you have learned -- what valid and invalid argument forms you are aware of, whether you have done proof by contradiction, and so on.)
 

promitheus

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Aug 19, 2017
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I think I'm beginning to see this a little clearer.

So p is the statement that 10=20 and if q is the statement that (-5)2=52

So p->q, but the argument is invalid as we are incorrectly using the converse of this implication.

I have done proof by contradiction and contrapositive implication. My course provides very few resources however ( I study via distance) and so have found this topic a bit hard to grasp. I have found some good video resources on youtube, but I must admit I have a lot more work to do!
 

Dr.Peterson

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Yes, some books call this the "converse error" or the "converse fallacy". Good work.

You could also say that the argument looks superficially like a proof by contradiction, but there is no contradiction!
 

promitheus

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Aug 19, 2017
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Thanks so much for your help. I have spent over 2 hours trying to figure this one out. I think it's time for a Sunday afternoon beer. :)
 

HallsofIvy

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You might be confusing this with "indirect proof" or "proof by contradiction". "Indirect proof" starts by assuming that the conclusion is false and using that to so show that the hypothesis is also false.
That works because it can be show that "if p then q" is equivalent to "if not p then not q". One argument is valid if and only if the other is. You are trying to say "if q then p" which is NOT a valid argument.
 
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