Show that 7-sq. root 5 is …
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Sia ? 6 years, 7 months ago
Let us assume that {tex}7 - \sqrt 5{/tex} is a rational number.
Then, there must exist positive co primes a and b such that
{tex}\begin{array}{l}7 - \sqrt 5=\frac{\mathrm a}{\mathrm b}\\\end{array}{/tex}
{tex}-\sqrt5=\frac{\mathrm a}{\mathrm b}-7{/tex}
{tex}\begin{array}{l}\sqrt5=7-\frac{\mathrm a}{\mathrm b}\\\end{array}{/tex}
{tex}\begin{array}{l}\sqrt 5=\frac{7\mathrm b-\mathrm a}{\mathrm a}\\\end{array}{/tex}
The right side {tex}\begin{array}{l}\frac{7\mathrm b-\mathrm a}{2\mathrm a}\\\end{array}{/tex} is a rational numbers so {tex}\sqrt5{/tex} is a rational number
This contradicts the fact that {tex}\sqrt 5{/tex} is an irrational number
Hence our assumption is incorrect and {tex}7 - \sqrt 5{/tex} is an irrational number.
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