Irrational numbers ( a ) , ( b ) exist that make ( a^b ) rational.
Prove with law of excluded middle and without fixing what ( a ) and ( b ) are.
Proof
Supporse ( x = sqrt{2} ^ sqrt{2} ) .
Case: ( x ) is Rational
( sqrt{2} ) is irrational, so ( a = b = sqrt{2} ) make ( a^b ) rational. ( ( sqrt{2} ) が無理数であることの証明は 証明: ( sqrt{2} ) は無理数 をご覧ください。)
Case: ( x ) is Irrational
begin{eqnarray*}
sqrt{2} & = & (sqrt{2}^sqrt{2})^sqrt{2}
& = & sqrt{2}^{sqrt{2} cdot sqrt{2}}
& = & sqrt{2}^2
& = & 2
end{eqnarray*}
Then, ( a = x = sqrt{2}^sqrt{2} ) and ( b = sqrt{2} ) are irrational and make ( a^b ) rational.
Above all, there are irrational numbers ( a ) and ( b ) that make ( a^b ) rational.
Law of Excluded Middle
形式論理学の用語で、ある命題についてその肯定と否定とがある場合、一方が真ならば他方は偽、他方が真ならば一方は偽であり、その両方のどちらでもない状態は存在しないというもの。 命題 ( P ) について ( P lor lnot P ) が必ず真となることをいう。
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I prove that infinite decimal of rational number is recurring decimal. It is decimal of rational number but is not finite decimal.
While the process of division may seem intuitive, I will provide a more concrete explanation as a formal proof.
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For example, when 89 is sequentially divided by 13, including decimal places, the remainders are as follows:
\begin{eqnarray*}
89 \div 13 & = & 6 \; \textrm{remainder is} 11 \\
110 \div 13 & = & 8 \; \textrm{remainder is} 6 \\
60 \div 13 & = & 4 \; \textrm{remainder is} 8 \\
80 \div 13 & = & 6 \; \textrm{remainder is} 2 \\
20 \div 13 & = & 1 \; \textrm{remainder is} 7 \\
70 \div 13 & = & 6 \; \textrm{remainder is} 5 \\
50 \div 13 & = & 3 \; \textrm{remainder is} 11 \\
110 \div 13 & = & 8 \; \textrm{remainder is} 6
\end{eqnarray*}
On the 7th division, the same remainder as in the 1st division appeared again. Since there are only 12 possible remainers when divided by 13, ranging from 1 to 12, if the division doesn’t result in a perfect quotient after 13 divisions, it becomes evident that at some point, the same remainder would recur, leading to a repeating decimal.
Continue reading Proof: Infinite Decimal of Rational Number is Recurring Decimal →
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Let’s assume the definition of rational numbers is known. Real numbers that are not rational are called irrational numbers.
Here, we will prove that \( \sqrt{2} \) is an irrational number.
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