Question

prove √5 is an irrational number ​

Answer 1

Answer:

Given: √5

Given: √5We need to prove that √5 is irrationalProof:Let us assume that √5 is a rational number.

Given: √5We need to prove that √5 is irrationalProof:Let us assume that √5 is a rational number.Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

Given: √5We need to prove that √5 is irrationalProof:Let us assume that √5 is a rational number.Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0⇒√5=p/q

On squaring both the sides we get,

$5 = {p}^{2} / {q}^{2} \\ ⇒5q²=p² ——(i) \\ p²/5= q²$

So 5 divides p

So 5 divides pp is a multiple of 5

$⇒p=5m \\ </em></p><p><em>[tex]⇒p=5m \\ ⇒p²=25m² ——(ii)$

From equations (i) and (ii), we get,

$5q²=25m²</em></p><p><em>[tex]5q²=25m²⇒q²=5m²</em></p><p><em>[tex]5q²=25m²⇒q²=5m²⇒q² \: is \: a \: multiple \: of \: 5</em></p><p><em>[tex]5q²=25m²⇒q²=5m²⇒q² \: is \: a \: multiple \: of \: 5⇒q \: is \: a \: multiple \: of \: 5$

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number√5 is an irrational number

Step-by-step explanation:

HOPE THIS HELPS PLEASE FOLLOW

Answer 2

$\texttt{If, \; possible, \; let} \; \mathtt{√11} \; \texttt{be \; rational}$

$\texttt{Then,}$

$\mathtt{√5 = \large \frac{a}{b}} \;\;\;\;\; [\texttt{where\; a \; \& \; b \; are \; co-primes} \\ \texttt{ \& \; b} \; \mathtt{≠ 0}]$

$\texttt{On \; squaring \; both \; sides,}$

$\implies \mathtt{5 = \large\frac{a²}{b²}}$

$\implies \mathtt{5b² = a² \;\;\;\; ----(i)}$

$\implies \mathtt{5} \; \texttt{divides} \; \mathtt{a²} \;\;\;\; [\mathtt{∵ \; 5} \; \texttt{divides} \; \mathtt{5b²}]$

$\implies \mathtt{5} \; \texttt{divides \; a}$

$\texttt{Let} \; \mathtt{a = 5c} \; \texttt{for \; some \; integer \; c.}$

$\texttt{Putting} \; \mathtt{a = 5c} \; \texttt{in \; (i), \; we \; get}$

$\;\;\; \mathtt{5b² = 25c²}$

$\implies \mathtt{b² = 5c²}$

$\implies \mathtt{5} \; \texttt{divides} \; \mathtt{b²}\;\;\;\; [\mathtt{∵ \; 5} \; \texttt{divides} \; \mathtt{5c²}]$

$\implies \mathtt{5} \; \texttt{divides \; b}$

$\texttt{Thus,} \; \mathtt{5} \; \texttt{is \; a \; common \; factor \; of \; a \& b.}$

$\texttt{But, \; this \; contadicts \; the \; fact} \\ \texttt{that \; a \; \& \; are \; co-primes, \; i.e. \; they} \\ \texttt{have \; no \; common \; factor \; other \; than} \; \mathtt {1.}$

$\texttt{The \; arises \; by \; assuming \; that} \; \mathtt{√5} \\ \texttt{is \; rational.}$

$\texttt{Hence,} \; \mathtt{√5} \; \texttt{is \; irrational.}$