Is this proposition $\sqrt{2}$ + $\sqrt{3}$ is an irrational number." true or false?

Keenan Santos
2022-07-08
Answered

Is this proposition $\sqrt{2}$ + $\sqrt{3}$ is an irrational number." true or false?

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jugf5

Answered 2022-07-09
Author has **18** answers

If it were rational, its square would also be rational, so $2+3+2\sqrt{6}$ would be rational and hence $\sqrt{6}$ would be rational.

therightwomanwf

Answered 2022-07-10
Author has **4** answers

$\sqrt{2}+\sqrt{3}=\frac{p}{q}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2={(\frac{p}{q}-\sqrt{3})}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{3+\frac{{p}^{2}}{{q}^{2}}-2}{2\frac{p}{q}}=\sqrt{3}$

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Let ${x}_{1},\dots ,{x}_{n}$ be positive rational numbers. If $\sqrt[{l}_{1}]{{x}_{1}},\dots ,\sqrt[{l}_{n}]{{x}_{n}}$ are all irrational numbers (where ${l}_{1},{l}_{2},\dots ,{l}_{n}\in {\mathbb{N}}^{\ast}$), does it follow that

$\sqrt[{l}_{1}]{{x}_{1}}+\cdots +\sqrt[{l}_{n}]{{x}_{n}}$

is an irrational number, too?

$\sqrt[{l}_{1}]{{x}_{1}}+\cdots +\sqrt[{l}_{n}]{{x}_{n}}$

is an irrational number, too?

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