Prove that the real number
√
2 +
√
5 is not rational.​

2 thoughts on “Prove that the real number <br />√<br /> 2 +<br /> √<br /> 5 is not rational.​”

1. Answer:

   In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. The adjective real in this context was introduced in the 17th century by René Descartes, who distinguished between real and imaginary roots of polynomials

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2. Step-by-step explanation:

   To prove Root2+root5 is irrational.

   proof:-

   First we need to prove root2 as irrational

   Let us prove it by contradiction method

   assume that root2 is rational

   we know that any rational no can be written in the form of p/q ,where p,q belongs to integer and q shouldn’t equals to 0 and HCF(p, q)=1

   so let us write p/q=root2

   p=root2(q)

   p^2=2q^2 . . . . .(1)

   we can observe in above equation that p^2 has two factor 2 and q^2

   If 2 is factor of p^2 then it is also the factor of p. . . . . .(2)

   let p=2m for some natural no. ‘m’

   then p^2=4m^2 . . .. . .. .. put this in (1)

   4m^2=2q^2

   q^2=2m^2

   similarly q^2 has factor 2 and m^2

   if q^2 has factor 2 then q also has factor 2 . .. . . .(3)

   from (2) and (3) we conclude that p and q both has factor 2

   but we have given that HCF of p, q should be 1

   so our assumption was wrong that root2 is rational

   Hence root2 is irrational .. . . . .(4)

   now we have to prove root2+root5 is irrational

   In same way

   p/q=root2+root5

   (p/q-root2)^2=(root5)^2

   (p^2/q^2)+2-2*p/q*root2=5

   (p^2/q^2)-3=p/q *root2

   ((p^2/q^2)-3)q/p=root2

   In LHS =rational

   but in RHS=irrational (root2 is irrational we have proved above (4)

   hence it also contradicts that root5 is rational so our assumption was wrong

   Hence root2+root5 is irrational

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