## ចំលើយលំហាត់ ខែ​មេសា 21, 2009

Posted by psvjupiter in គណិតវិទ្យា, អូឡាំព្យាដ.

ខាងក្រោមនេះជាសំរាយលំហាត់ខ្ញុំថ្ងៃមុន ខ្ញុំទើបតែមានពេលទំនេរទើបផូសយូរបន្តិច។

ចំពោះលំហាត់ទី៣​ ខ្ញុំគិតចេញហើយបុន្តែ ខ្ញុំប្រើស្វ័យគុណចំនុចនិងទ្រឹស្តីបទស៊ីនុសក្នុងត្រីកោន ខ្ញុំគិតថាប្រហែលជាមានការលំបាកសំរាប់អ្នកទាំងអស់គ្នា ក្នុងការយល់សំរាយខ្ញុំក៏សំរេចចិត្តមិនដាក់នៅលើប្រកាស
ប្រហែលលោកគ្រូវិចិត្រ​មានរបៀបងាយជាងខ្ញុំ។ សង្ឈឹមលោកគ្រូបញ្ចេញថ្វៃដៃម្តង។ 😀

## មតិ»

1. Mony - ខែ​មេសា 23, 2009

Hello, I’ve one nice exercise for you

If a=b (a,b>0)
solve that: 1=2

2. kienforcefidele - ខែ​មេសា 23, 2009

$a=b$
So $a^{2}=ab$
$a^{2}-b^{2}=ab-b^{2}$
$(a-b)(a+b)=b(a-b)$
$a+b=b$
But a=b; so:
$b+b=a$
$2b=a$
$2b=b$
=> $2=1$
However, this is wrong. See that $a=b$ so $a-b=0$ and we can’t divide $0$ by $0$ because it’s indefinite.

3. Mony - ខែ​មេសា 23, 2009

wow great! You’re right!

4. jupiter - ខែ​មេសា 23, 2009

awe some! I appreciate that. I accept that solution 😀 I can’t think like that lolz… of course it is correct solution because false imply false….. (when you learn logic you will understand, false problem can give false solution).

5. Vatey - ខែ​មេសា 24, 2009

In one party, there are N people. Show that: At least 2 people greet with the same amount of people in the party!

Sorry I can’t write it in Khmer cuz I don’t know how to!

6. jupiter - ខែ​មេសា 24, 2009

It is of type pigeon hole principle…
There are N people in party, for one person the number of people he can greet would rank from 0 to N-1.

if it is 0 that mean that person didnot greet any one. Suppose there are X people who didnot greet anyone. So the number of people that involve in greeting at the party may reduce to N-X and number each person in that {(N-x)}group can greet range from 1to N-X-1. by pigeon hole principle at least two pp have same greeter…

what is pigeon hole principle? It is the siimplest principle. Now suppose you have 10 room, and 11 people are going to stay in all rooms, then at least one room have >=2people. U can not place 11 stuff one by one in 10 stuff lolz, so one must have at least 2 people.
As in your case you may take N=23, X=10, then N-x=13, N-x-1=12. It is like trying to put 13 people in 12 rooms. ::D

7. jupiter - ខែ​មេសា 24, 2009

There is a slight mistake in my proof above, what if X=N? then X-N-1 is negative!
Actaully X can only be 1
why? If X>1 that mean X>=2 and we are done! because 2 people have the same greet number and it is 0…

8. Vatey - ខែ​មេសា 24, 2009

Can you tell me what grade is this exercise?

9. Vatey - ខែ​មេសា 24, 2009

Anyone could tell me, how to write in khmer please!

10. ពិសិដ្ឋ - ខែ​មេសា 24, 2009

11. Vatey - ខែ​មេសា 28, 2009

Thanks I can read all of the Khmer words in this blog properly but I can’t type it in Khmer! So I’ve to type it in Microsoft word and paste it in here, isn’t it?

ពិសិដ្ឋ - ខែ​មេសា 28, 2009

You can do like that or write it directly. That’s OK! 😀

12. Vatey - ខែ​មេសា 29, 2009

thanks but i really can’t write in here directly or you can say because i don’t know how to! 🙂

ពិសិដ្ឋ - ខែ​មេសា 29, 2009

hrmm… Maybe you don’t know how to type Khmer Unicode. First you need to install Khmer Unicode keyboard.