Question

prove that there is no rational no whose square is 2

Answer 1

[Tex]\bold{hi}[/tex]
Let the number be. x

Acc to question,
x^2 = 2
x= √2

but root 2 is irrational so x is also irrational

Answer 2

$Let\ there\ is\ a\ rational\ number\ \frac{a}{b}\ whose\ square\ is\ 2. \\ \\ Here,\ a\ and\ b\ have\ no\ common\ factors\ except\ 1 \\ and\ b\ is\ not\ equal\ to\ 0. \\ \\ And\ \frac{a}{b}\ is\ of\ simplest\ form. \\ \\ (\frac{a}{b})^2 = 2 \\ \\ = \frac{a^2}{b^2} = 2 \\ \\ \\ a^2 = 2b^2 \\ \\ Here\ we\ can\ find\ that\ a^2\ is\ an\ even\ number\ because\ a^2\ is\ 2 \\ multiplied\ by\ b^2. \\ \\ \therefore a\ is\ an\ even\ number\ because\ only\ even\ numbers \\ have\ squares\ even.$

$Let\ a = 2m \\ \\ 2 = \frac{a^2}{b^2} = \frac{(2m)^2}{b^2} = \frac{4m^2}{b^2} \\ \\ \\ 2b^2 = 4m^2 \\ \\ b^2 = 2m^2 \\ \\ Here\ we\ can\ find\ that\ b^2\ is\ an\ even\ number. \\ \\ \therefore b\ is\ an\ even\ number. \\ \\ It\ seems\ that\ a\ and\ b\ have\ a\ common\ factor\ 2 \\ and\ \frac{a}{b}\ is\ not\ the\ simplest\ form. \\ \\ This\ is\ a\ contradiction. \\ \\ \\ \therefore There\ is\ no\ rational\ whose\ square\ is\ 2.$

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