Question

Question number 19 and 20 guys please help
Maths genius

Answer 1

         

$\sf{19. Assume that} \ 3+5\sqrt{2}\ \sf{is rational.} \\ \\ Let} \ 3+5\sqrt{2}=x \\ \\ \sf{If both sides of} \ \ 3+5\sqrt{2}=x\ \sf{are rational, then so will be} \ \ \frac{x-3}{5}=\sqrt{2}. \\ \\ \\ x=3+5\sqrt{2} \\ \\ x-3=5\sqrt{2} \\ \\ \frac{x-3}{5}=\sqrt{2} \\ \\ \\ \sf{But} \ \sqrt{2}\ \sf{is irrational. \\ \\ Thus this contradicts our assumption. \\ \\ \\ \therefore\ 3+5\sqrt{2}\ \sf{is irrational.}$

$\sf{20. Assume that} \ 2\sqrt{3}-1\ \sf{is rational.} \\ \\ Let} \ \ 2\sqrt{3}-1=x \\ \\ \sf{If both sides of} \ \ 2\sqrt{3}-1=x\ \sf{are rational, then so will be} \ \ \frac{x+1}{2}=\sqrt{3}. \\ \\ \\ x=2\sqrt{3}-1 \\ \\ x+1=2\sqrt{3} \\ \\ \frac{x+1}{2}=\sqrt{3} \\ \\ \\ \sf{But} \ \sqrt{3}\ \sf{is irrational. \\ \\ Thus this contradicts our assumption. \\ \\ \\ \therefore\ 2\sqrt{3}-1\ \sf{is irrational.}$

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