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MTH202 Assignment No 02 Solution & Discussion Due Date:24-01-2017

Question: 01   Prove by contradiction that     is irrational.

 

Question# 02 (a)  Find the number of words that can be formed from the word VIRTUAL if the words end with letter  T and start with letter L. (Words need not be in the dictionary and each letter should be used once)

(b) How many 4-digit even numbers can be formed using digits 1,2,3,5 using each digit once? 

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Question no.1

Prove by contradiction that     is irrational.

 

Solution:

Suppose 7+11√2 is rational

Then by definition of rational

7+11√2=a∕b

For some integers a and b with b≠0.

Now consider

11√2=a∕b-7

11√2=a-7b/b

√2=a-7b/11b

Since a and b are integers so a-7b and 11b and 11b≠0;

Hence √2 is a quotient of the two integers a-7b and 11b with 11b≠0.

Accordingly , √2 is rational(by def. of rational).

This contradicts the fact because √2 is irrational. Hence our supposition is false and so 7=11√2 is irrational.

 

ye to easy hy Q no2 ka baten keya ans ata hy mera to (a) ka ans ata hy 5!=120 ans  (b) ka ans hy 4.3.2=24 keya ye ans tik hy dono k 

Aamir Shehzad  thanks for sharing 

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mth202 assignment #2 idea solution 2017

Mth 202 Assignment
Question no.1
Prove by contradiction that     is irrational.
 
Solution:
Suppose 7+11√2 is rational
Then by definition of rational
7+11√2=a∕b
For some integers a and b with b≠0.
Now consider
11√2=a∕b-7
11√2=a-7b/b
√2=a-7b/11b
Since a and b are integers so a-7b and 11b and 11b≠0;
Hence √2 is a quotient of the two integers a-7b and 11b with 11b≠0.
Accordingly , √2 is rational(by def. of rational).
This contradicts the fact because √2 is irrational. Hence our supposition is false and so 7=11√2 is irrational.

Q2(A)
5!=5.4.3.2.1=120
(b) =3!
3.2.1=6 ans

mth202 assignment #2 idea solution 2017

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Q2 (B) may 4 digits hai tu kia yeh asy slove ni hoga
5*3*1*2=30
last digit ko even rahkna hai na

https://www.youtube.com/watch?v=Dw231YQi9yE

Q1 Ka yeh ans ata hai
√2=7b-a/11b as par hand out

But here we have to prove 7+11√2

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