how to show that √ 2 + √ 5is irrational  

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jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Before solving this you must know that;

  • The square of an odd number is odd
  • The square of an even number is even.
  • If square of a number is even that means it is divisible by 2.


Lets assume sqrt(2) + sqrt(5) is rational.

Let y = sqrt(2) + sqrt(5) for any rational number y.

Then  y^2 = (sqrt(2) + sqrt(5))^2

        y^2 = 2+5+2*sqrt(2)*sqrt(5)

       y^2  = 2+5+2*sqrt(2*5)

      y^2  = 2+5+2*sqrt(10)


sqrt(10) = (y^2-7)/2

Since y is rational sqrt(10) = (y^2-7)/2 is also rational.

Therefore sqrt(10) is rational.

So we can write sqrt(10) = p/q where p,q are relatively prime.

sqrt(10) = p/q

         10= p^2/q^2

 10*q^2= p^2---------------(1)

Left hand side is an even number which is divisible by 2. Therefore right hand side should also be even and thus divisible by 2. Then p is divisible by 2.

So we can write p = 2*k where k is a integer.

Then from (1) we get ;

10*q^2 = (2k)^2

10*q^2 = 4*k^2

5*q^2  = 2*k^2

Right hand side of the equation is even and divisible by 2. Then left hand side should also be divisible by 2. Since 5 is a odd prime number q^2 should be even. Thus q should be even. Then q is divisible by 2.

So finally p and q are not relatively prime integers. Our assumption is wrong.

Then sqrt(10) is not rational. Hence y is not rational. That means sqrt(2) + sqrt(5) is not rational.



vaaruni's profile pic

vaaruni | High School Teacher | (Level 1) Salutatorian

Posted on

Let  sqrt(2) + sqrt(5) = x

( sqrt(2) + sqrt(5) )^2 = x^2   [ squaring both sides ]

Or,  (sqrt(2))^2 + 2* sqrt(2) * sqrt(5) + (sqrt(5))^2 = x^2

Or,  4 + 2*sqrt(2) * sqrt(5) + 5 = x^2

Or, 9 + 2* sqrt(10) = x^2

in the expression -> 9 + 2 sqrt(10)  Since,  sqrt(10) is ir-rational therefore 2*sqrt(10) is ir-rational and  9 + sqrt(10) is also ir-rational. Hence x^2  is ir-rational and  x is also ir-rational.                         Therefore,  sqrt(2) + sqrt(5)  is ir-rational    <---- Answer  



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