Thanks, I think I get it now. I have shown the equation only holds if c has that particular value. If it has any other value, it doesn't hold. So in general, no, it is not right to assert that

a + (bc) = (a + b)(a + b)



I think the author was referring to identity. Cheers.



Thanks.



Not sure what this means. How to interpret ≡?

Hello Sophisticat, thanks for your reply. I have however decided to let the matter rest. I'm getting on fine with my mathematics book, and this is clearly a bit beyond my current capabilities.

Thanks to all who contributed to the thread.

@fdrake,

Thanks for clearing that up for me. So what I don't understand is how one can suddenly grow broader to any number and still use that broadest definition for our to be proven theorema, like andrewk wrote:

d0 + d1(10^1) + d2(10^2) +....+ dn(10^n)

So we can just take this and generalize our findings to represent any number? Why? It's the +....+ dn(10^n) I have trouble with. And how can I know what to do in cases like this? I.e,. using this generalized formula of a number to proof our statement.

@andrewk, Hrvoje,

I must admit that this is all a bit beyond me.

Let's see where I lose track.

I understand that any number can be rewritten as

d0 + d1(10) + d2(10^2) + ad infinitum

Where d stands for a digit between 0 and 9. ( I also don't know how to write the make-up you used)

So, what we are trying to proof is that:

if d0 + d1(10) + d2(10^2) / 3

then

d0 + d1 + d3 / 3

Do I at least understand the theorem to be proved?

First small step I lose track with: how do you know when you can write any number as

d0 + d1(10) + d2(10^2) + dn(10^n) ? I have trouble understanding this method of notation and when to use it and what it means.

@TheMadFool, Sophisticat, AndrewK

My head is still spinning a bit. I'm still at TheMadFool's proof of

- for any number divisible by 3, the sum of the digits is divisible by 3

Just to be sure, k and r each stand for a digit in any two digit number?

So if a number is divisible by 3, the sum of k and r should be divisible by 3?

Hello TheMadFool,

Thanks for your reply.

Have we now converted any two digit number into

3k + (k + r)/3

Where the left stands for the first digit, and the (k+r)/3 stands for the second digit?

Hello fdrake, andrewk and Tim3003,

Thanks for your input and continuation of the discussion.

To be honest, I'm lost at

- for any number divisible by 3, the sum of the digits is divisible by 3

I wouldn't know how to even begin proving this.

I scribbled around on paper a bit and thought, well, since the number is divisible by 3, it can also be written as 3n.

So we have a number 3n, if you write it down in the original form, before the division by 3, then the sum of the digits is also 3n, or say 3m. How would one express 'sum of digits' more formally in order to reason with it mathematically?

Hello fdrake,

Thank you for your reply and additional challenge.

Lets see.

I want to prove:

P2: The last digit of any number determines it being even or uneven.

I can use this fact about numbers:

Any number can be written in the form 10 * k + r, where r is the last digit in their decimal notation.

I'll make use of this axiom:

An even number + an even number is an even number, or:

2m + 2n = 2(m+n)

And, for any k, 10 * k is even, for

10 * k = 2(5k), thus containing 2 as a factor, making it even.

Then, if r is an even last digit number, the options for the last digit are 0, 2, 4, 6, or 8. All even numbers because of the 2m definition.

If we add any even r to 10 * k, then the result will be even, for we are adding an even number to an even number. Thus, if the last digit is even, the whole number is even.



To prove the same for uneven numbers:

An even number + an uneven number is an uneven number, or

2m + 2n+1 = 2(m+n)+1

And, for any k, 10 * k is even, for

10 * k = 2(5k), thus containing 2 as a factor, making it even.

Then, if r is an uneven last digit number, the options for the last digit are 1, 3, 5, 7, or 9. All uneven numbers because of the 2n+1 definition.

If we add an uneven r to 10 * k, then the result will be uneven, for we are adding an uneven number to an even number. Thus, if the last digit is uneven, the whole number is uneven.


I guess I'm using too much language, I am not used to formal reasoning yet. I enjoy it, though.

Hello fdrake,

Thanks for your reply. Happy to read that my proof, though cruder, works. Sorry for my late response, lots of other responsibilities, you know the deal.

Now for your question:

Can I proof:

''If a number ends in 0, 2, 4, 6, or 8, then it is even.''

I would go about it this way:

P1: An even number is defined by it being divisible into two equal parts without any leftovers, aka, two equal integers.

P2: The last digit of any number determines it being even or uneven.

P3: Therefore we only have to take into account the last digit to cover all numbers.

P4: 0, 2, 4, 6 and 8 are divisible into two equal integers

P5: the other options for number endings, 1, 3, 5, 7, and 9 are not

Conclusion: any number ending in 0, 2, 4, 6 and 8 is even.

Problems with this proof:

*P1 is not proven
*P2 is not proven
*In general: using too many words instead of symbolism

Cheers!

Hello fdrake & unenlightened,

Thank you for your input. I think I understand, or at least I understand it better now, despite some conflicting instructions :wink:

@Sophisticat

Thank you for your reply. Still not sure how one can make a deduction like this:

All people wear blue jackets

therefore

Some people wear blue jackets

However if we start with the conclusion here, and take it as a premise or a statement, then I do see that it means or could mean that 'there is at least one person that is wearing a blue jacket'. Though to be perfectly honest, in natural speech, when I use 'some', I think of more than one person, say a group of 10 people, and when I say 'some are wearing blue jackets', I would mean like 2 or 3, not just one. If it was just the one, I would have said: 'There is one person wearing a blue jacket'.

Long story short, should I just take it as a rule or convention when I encounter in Logic the statement 'All A's are B's', you can validly deduce 'Some A's are B's', despite natural speech?

Thank you Pair o'Ducks, fdrake and Sophisticat!

I had other duties the last few days hence my late reply. It sure is reassuring that the reasoning is actually invalid, though the book says it is valid. I found it to be invalid precisely because of reasons mentioned in this thread. The book probably did mean integers, if integer implies : 'there are no other numbers other than even or odd'. In that case, if I'm not mistaken, the total number of possible even numbers is equal to the possible number of even numbers squared, hence it is not possible to not find a number that is even that does not have a square that is also even, which is a long way of saying:

If a number is even, if you square it, it will be even
and
If you have an even squared number, if you take the root of it, it will be even

Regarding SophistiCat's concern of focusing only on the 2nd premise: I did use the 1st premise in my notes, which I didn't write out all here.

In any case thanks again and I am now ready to move on to the next exercise!

Hello Unenlightened, thanks for your reply as well.

I've tried to apply your instructions, but I was still a bit confused. Is this what you had in mind? I interpreted the premise 'Americans curse' as 'All Americans curse'.



In this second drawing I interpreted the premise as 'Some Americans curse'

Hello SophistiCat, thanks again for your reply. What I gather from it is that, indeed, there is some vagueness in the premises. What I fail to understand however is that you can conclude 'Some Americans curse' from the premise 'All Americans curse'. You see, I thought if one says 'Some people are nice', one means that 'some people are nice, and some are not', rendering the deduction from 'All people are nice' impossible. Basically, why is this valid:

All A's are B's

Therefore

Some A's are B's

While I take the 2nd premise to mean: Some A's are B's and some are not. Since some A's are not B's, the conclusion cannot follow from the premise.

Sorry if this makes it all too complicated. But it's a basic confusion that will return often in logic I have the intuition.

Cheers. When interpretation comes into play, I've learned I'm usually interpreting it wrongly. Common sense I find not obvious at all.

Would be cool if you could take a peak at the Venn-diagram I drew and the error in my reasoning there too. See the post above yours.

Remaining on the topic I found another tricky syllogism, which I approached by drawing a Venn-diagram, like you did.

This is the argument:

No decent people curse
Americans curse

Therefore

Some Americans are not decent

The book says this conclusion is valid. I fail to see why. I answered invalid, because based on the premises it says 'all Americans are not-decent', not just 'Some Americans are not decent'.

Here's the Venn-diagram I drew, would be cool to know if I drew it correctly or not, I feel like if I can become good at drawing these diagrams I can finally understand logic.

[img]http://

Hello SophistiCat,

Thanks for your reply.

Yes, you are right, I interpreted 'are' as 'being equal to', not as class membership. Is thinking in class memberships the right way to approach syllogisms in general? I find it hard to determine what context I can, and cannot use, since, as I understand it now, in logic, we ignore anything that's not in the premises.

But how about this famous argument from Aristotle, where he says:

If A is equal to B, and B is equal to C, then A is equal to C.

In this case we have to interpret 'equal to' as identical to, not as class membership?

If A = B and C = B, then A = C would be correct in that case.

Hello MadFool,

Thanks for your reply. I am not familiar with the technical jargon of distribution. But yes, I can see that being valuable is the common ground. Not sure how that makes the reasoning in my 1st post invalid.