Saved Bookmarks
| 1. |
Prove that root two is an irrational numbet |
| Answer» If possible let√2 be the rational no. In The form of. a/b where a and b are Co prime no.Now ,√2= a/b.Squaring both side.(√2)2 = a2 / b2..2 = a2/b2...b2= a2/2 ...So, 2a is divisible by 2 so, a is divisible by 2(by theorm). Now let a=2cSo . b2 =( 2c)2/2 ...b2= 4c2/2. ..b2= 2c2...b2/2= c2..Hence, ...B2 is divisible by 2 so, B is divisible by 2(by theorm)....Hence our contradiction is wrong and√2 is irrational ....proved.... Here, a2 = a square... | |