Question

prove that root 3 minus root 2 is an irrational number

pls answer with full process ​

Answer 1

Answer:

0.32

Step-by-step explanation:

√2=1.41

√3=1.73

.°. √3-√2= 1.73 - 1.41

=0.32

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Answer 2

GIVEN:

\$A \:numbrer\:3-\sqrt{2}$

TO PROVE:

→It is irrational.

PROOF:-

On the contrary let us assume that it is a rational number .

So it can be expressed in the form of $\dfrac{p}{q}$ where p and q are integers and q≠0 and p and q are co-primes.

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SO, ATQ,

=>$\dfrac{p}{q}= 3-\sqrt{2}$

On squaring both sides,

=>$(\dfrac{p}{q}) ^{2}=(3-\sqrt{2}) ^{2}$

=>$\dfrac{p^{2}}{q^{2}}=(3) ^{2}+\sqrt{2}^{2}-2×3×\sqrt{2}$

$\large\green{\boxed{(a-b) ^{2}=a^{2}+b^{2}-2ab}}$

=>$\dfrac{p^{2}}{q^{2}}=9+2-6\sqrt{2}$

=>$\dfrac{p^{2}}{q^{2}}= 11-6\sqrt{2}$

=>$11-\dfrac{p^{2}}{q^{2}}=6\sqrt{2}$

=>$\dfrac{11q^{2}-p^{2}}{q^{2}}=6\sqrt{2}$

=>$\dfrac{11q^{2}-p^{2}}{6q^{2}}=\sqrt{2}$

But, now we have arrived at a contradiction,

LHS is a rational number whereas RHS is a irrational number.

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So, our assumption was wrong that $3-\sqrt{2}$ is a rational number.

Hence $3-\sqrt{2}$ is a irrational number.

$\huge\orange{\boxed{Hence\:proved}}$

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