Question

Prove√7 -2√3 is
irrational​

Answer 1

Answer:

assume that √7-2√3 is rational

√7-2√3=a/b

therefore,a/b is rational

in this √7-2√3 is irrational

Step-by-step explanation:

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Answer 2

$\huge{ \underline{ \mathfrak{ \green{ \: Answer}}}}$

☆In order to prove $\sqrt{7}$ is irrational , I will contradict that it is rational.

→Let $\sqrt{7}$ - 2$\sqrt{3}$ is rational

→Then it can be written in the form of $\frac{p}{q}$ where p and q are positive integers and q≠0

$\implies$$\sqrt{7}$ - $2{\sqrt{3}}$ = $\frac{p}{q}$

→Squaring both sides

$\implies$(√7 - 2√3)² = ${\frac{p}{q}}^{2}$

$\implies$ 7 – 4√21 + 12 = ${\frac{p}{q}}^{2}$

$\implies$ 19 - 4√21 = ${\frac{p}{q}}^{2}$

$\implies$ -4√21 = ${\frac{p}{q}}^{2}$ -19

$\implies$ √21 = $\frac{19q^2- p^2}{4q^2}$

→We can see that LHS is rational , because p and q are integers and q≠0. This means RHS is also rational

→But it contradicts the fact that √21 is irrational. This contradiction has arised due to our wrong assumption

→So our assumption is wrong and thus √7-2√3 is irrational