|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Jul 2008 21:28:49 IST
|
|
|
prove that cube root of 2 is irrational
|
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
16 Jul 2008 21:31:24 IST
|
|
|
OK posting it. It is very easy.
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
16 Jul 2008 21:37:19 IST
|
|
|
1. suppose is rational so it is= p/q where p and q are integers with no common divisor.
2. raise to exponent of 3 and we have 2 = (p/q)3 or 2q3=p3
3. in LHS we have an even number , so in RHS q must be even. let say p= 2 r where r is an integer.
4. Substitute in 2 q^3=p^3 we have 2 q3=(2r)3 or
2 q3=8 (r3) or if we divide by 2, q3=4 (r3).
5. Now in the RHS we have an even number, so the LHS must be even or q = 2 s, where s is an integer.
6. From the last relation (q = 2 s) and p= 2 r (obtained above), we conclude that q and p have 2 as a common divisor.
7. The steps 1 and 6 are contradictory.
|
this reply: 7 points
(with 1
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
16 Jul 2008 21:40:02 IST
|
|
|
![\text{ Let cube root of 2 is rational}\\\\Then\ \sqrt[3]{2}=\frac{p}{q}\ \text{ (where (p,q)=1 and q is not equal to 0) }\\\\\text{Cubing both sides}\\\\2=\frac{p^3}{q^3}\\\\\Rightarrow p^3=2q^3\\\\let\ p=2k\\\\8k^3=2q^3\\\\Or\ q^3=2(2k^3)\\\\\text{ So 2 is a common factor for both p and q}\\\\\text{ Hence our assumption is wrong as (p,q)=2 and not (p,q)=1}\\\\\text{Hence}\ \sqrt[3]{2}\ \text{is irrational}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/4/9/44962636a861a42cf151308094565057aa69eb60.gif)
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
16 Jul 2008 21:53:32 IST
|
|
|
thx
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
