Math, asked by jamesstephen2015, 8 months ago

how to prove √2 is irrational

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Answered by XxMissPaglixX

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$let \: \sqrt{2} \: be \: rational \\$

So,

$\sqrt{2} = \frac{p}{q} \\ {( \sqrt{2}) }^{2} = { (\frac{p}{q}) }^{2} \\ 2 = \frac{ {p}^{2} }{ {q}^{2} } \\ 2 {q}^{2} = {p}^{2} - eq \: 1 \\$

Let,

$p = 2m \\ {p}^{2} = {(2m)}^{2} \\ {p}^{2} = 4 {m}^{2} \\ 2{q}^{2} = 4 {m}^{2} (from \: eq \: 1) \\ {q}^{2} = \frac{4 {m}^{2} }{2} \\ {q}^{2} = 2 {m}^{2} - eq \: 2$

From eq 1 and 2 we have 2 as a common factor for both p and q which contradicts our supposition that root 2 is not a rational no.

$\sqrt{2} \: is \: a \: irrational \: no.$

Hence Proved

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Answered by harinichimbili

2=a^2/b^2

2b^2=a^2 2/a

2b^2=(2k)^2 a=2k

2b^2=4k^2

b^2=2k^2

2/b

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how to prove √2 is irrational​

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So,

Let,

From eq 1 and 2 we have 2 as a common factor for both p and q which contradicts our supposition that root 2 is not a rational no.

Hence Proved

how to prove √2 is irrational