Because this is not form in p/q

Let take √5 as a rational numberIf a and b are two co-prime number and b is not equal to 0.We can write √5 = a/bMultiply by b both side we getb√5 = aTo remove root, Squaring on both sides, we get5b2 = a2 ……………(1)Therefore, 5 divides a2 and according to a theorem of rational number, for any prime number p which is divided \’a2\’ then it will divide \’a\’ also.That means 5 will divide \’a\’. So we can writea = 5cand putting the value of a in equation (1) we get5b2 = (5c)25b2 = 25c2Divide by 25 we getb2/5 = c2again using the same theorem we get that b will divide by 5and we have already get that a is divided by 5but a and b are co-prime number. so it is contradicting.Hence √5 is an irrational number

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 = √5qwe know that \’p\’ is a rational number. so √5 q must be rational since it equals to pbut it doesnt occurs with √5 since its not an intezertherefore, p =/= √5qthis contradicts the fact that √5 is an irrational numberhence our assumption is wrong and √5 is an irrational number.\xa0

Let us consider that √5 is a “rational number”.We were told that the rational numbers will be in the “form” of \\frac {p}{q}form Where “p, q” are integers.So, \\sqrt { 5 } = \\frac {p}{q}p = \\sqrt { 5 } \\times qwe know that \’p\’ is a “rational number”. So 5 \\times q should be normal as it is equal to pBut it did not happens with √5 because it is “not an integer”Therefore, p ≠ √5qThis denies that √5 is an “irrational number”So, our consideration is false and √5 is an “irrational number”.
Let us assume√5 is irrational number.So, √5 is in the form of p/q where p and q are some Integers where q≠0.Since ,We divide numerator and denominator by common factor to get a and b where a and b are co -prime.So , √5=a/b √5b = aSquaring both side,(√5b)^2 =a^2
Prove that√5 is irrational

5 is not an irratonal. it rational number because we can write 5 in the form of\xa0{tex}p\\over q{/tex}