Question

Prove that √5 is irrational​

Answer 1

Answer:

Let us prove that √5 is an irrational number, by using the contradiction method. So, say that √5 is a rational number can be expressed in the form of pq, where q ≠0. So, let √5 equals pq. Where p, q are co-prime integers i.e. they do not have any common factor except '1'.

Answer 2

Question: Prove that √5 is irrational

Answer:

Let us assume $\sqrt{5}$ is rational no.

$\sf \: therefore \: \sqrt{5} = \frac{p}{q} \\ \sf \: where \: p \: and \: q \: are \: coprime \: factors \\ \sf \: and \: q \: \cancel{ = } \: 0$

By taking squares on both sides

${ \sqrt{5} }^{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ \boxed{ \sf \: 5 {q}^{2} = {p}^{2} }$

Now, p² is divisible by 5

therefore,p is divisible by 5

$p = 5r \\ \sf \: by \: taking \: squares \\ \: {p}^{2} = 25 {r}^{2} \\ \hookrightarrow \sf but \: {p}^{2} = 5 {q}^{2} \\ \sf \: by \: putting \: value \\ 5 {q}^{2} = 25 {r}^{2} \\ \boxed{ \sf {q}^{2} = 5 {r}^{2} }$

Now, q² is divisible by 5

therefore,q is divisible by 5

From above we get,

both p and q are divisible by 5

↪But it contradicts the fact that p and q are coprime factors.

Therefore our assumption was wrong

$\rightarrow \sf \: \sqrt{5} \: is \: irrational \: no.$

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