Answers

2015-05-12T20:59:14+05:30
Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are intezers.
so, √5 = p/q
     p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesnt occurs with √5 since its not an intezer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.
2 5 2
once see in our text books. they gave in this pattern only
i m having book published by "the Government Of Andhra Pradesh, Hyderabad"
published by EMESCO Books
yup mee too
do u stay in hyd
The Brainliest Answer!
2015-05-12T21:01:41+05:30
If possible, let us assume that , \sqrt{5}  is irrational.

 \sqrt{5} =a÷b
 \sqrt{5} ×b=a
b \sqrt{5} =a
squaring on both sides
(b
\sqrt{x})^2  = a^{2}
 a^{2} =5 b^{2}      this is first equation
 a^{2}  is a multiple of 5, a^{2}  is multiple of 5,a is also multiple of 5.
 a^{2} =(5 l^{2} )
 a^{2} 25 l^{2}    this is second equation
equate equation :- 1&2
25l=5 b^{2}
25l÷5= b^{2}
5l= b^{2}
 b^{2}  is a multiple of 5 and b is also multiple of 5
As , a and b are multiples of 2 which are contradictory. Our assumption is wrong.
 \sqrt{5}  is irrational

3 3 3