Basic Mathematics - 2019-09-21

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yh05

12:41

I get what you meant now. The first person who proves something would also have had guessed it and then proved it. (because there would not be anyone else to tell him that he is 100% correct)

I think I wasn't uncomfortable with the guess-first strategy but rather confused with the proof technique involved.

I found a better example of the method of proof in elementary geometry. https://imgur.com/h6RLP81
It is taken from the book Elementary Geometry by Edwin E. Moise.

In step 1, the author shows how to locate the points X based on the relationship between their distance from F and r, s.

user21820

12:57

@yh05 Yup to your first paragraph. Give me a while to take a look at your new example.

Yes I think you got it right. If you always follow strictly rigorous reasoning in a proof, then you will never get a false conclusion. However, we are free to pursue nonrigorous intuition in trying to guess true sentences before we attempt to prove them.

Often, we also try to guess relationships between properties; they help us understand the properties better. So if we want to understand property A (on some type of objects), we might want to look for a property B (on that same type of objects) that we can prove A and B to be equivalent (for those objects).

One way we can achieve this is to look for logical consequences of A, and try to figure out which consequences are strong enough to also imply A (i.e. the reverse implication).

Sometimes problem solving can also be helped by such a tactic. For example, if the problem tells us to prove that conditions A,B,C imply conclusion D, then we might try finding a different condition A' that is easier to work with than A, that we can prove is a consequence of A,B,C, and hope to also find a proof that A',B,C imply D. But how to choose A'?

In most good problems, where A,B,C are all necessary (i.e. D may not hold if one of them does not hold), we would need to figure out an A' that is not too weak compared to A, given both B,C.

And in many such situations, we end up finding an A' that is equivalent to A, though for the sake of solving the problem we won't need to prove the equivalence. But this may explain why we 'accidentally' get equivalences even if we are only looking for implications.

yuvraj singh

@user21820 hi

user21820

@yuvrajsingh Hi.

yuvraj singh

@user21820 have very small question

@user21820 On permutation and combinations, a problem is in party there are 5 Russian and 5 British couple, we have to find total no of shake hands if Russian women will. Not hand shake with any man, and any women will not shake hand with her own husband
What I do is total hand shakes are 190,i minus shakes hand between Russian women and British man that is 25 - shake hands of us Russian women with her own husband, - shake of British ladies with their husband am I right or I missed an case will anybody help me

user21820

13:13

@yuvrajsingh You subtracted "shakes hand between Russian women and British man", but what about Russian women with other men?

yuvraj singh

@user21820 there is no boundless for Russian ladies

user21820

@yuvrajsingh You wrote: "find total no of shake hands if Russian women will. Not hand shake with any man, and ..."...

yuvraj singh

@user21820 think it is misprint then my answer is correct

@user21820 how to write it using combination formula

user21820

@yuvrajsingh Well, for the question as you wrote it above (minus the grammar and punctuation errors), your answer is wrong. We can't talk about a question that is not written down.

yuvraj singh

@user21820 sir, I extremely sorry for punctuation and grammar

@user21820 one more doubt can i

user21820

13:23

@yuvrajsingh No problem, but how do you know I am a "sir"? =P

@yuvrajsingh Sure.

yuvraj singh

@user21820 I have issue understanding distribution of discrete thong among childern

Let say I have 9 different I phone, I want distribute among 4 friends, how many possible ways I can do this

user21820

@yuvrajsingh Just start distributing them one by one and count how many possible choices you can make at each step...

yuvraj singh

@user21820 sorry but are you kidding

user21820

@yuvrajsingh I am not! You can learn basic combinatorics only by doing this.

yuvraj singh

@user21820 I thought it is 8c4

user21820

13:30

If you are guessing, then you're learning in the wrong way. Follow what I told you to do and see if you can figure out the right way yourself, and be sure that you are right.

yuvraj singh

No sir I Don, t

user21820

If you cannot, then you can try the even simpler method of listing all the possible solutions systematically. The more systematic you are, the faster you will figure out the mathematics.

yuvraj singh

@user21820 I read the basics of combination

@user21820 I was thinking definition say when order matter you write 4^8 and order do not matter we write 8c4

yh05

14:02

@yuvrajsingh Take user21820's advice. You will see that in the proofs of (n^r) and (n choose r).

Btw, it's either 4^8 (with an order) 4c8 (no order is required) or 8^4, 8c4. Not what you stated.

1 hour later…

user21820

15:25

@yh05 Haha I did a double-take when I read your "4c8", but indeed. Lol.

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