グローバルビジネスで役立つ数学（72）無理数の証明（英語版）

Hi there!

It would be too much to call “irrational exuberance” that company employees beam excitedly before a long vacation, even without specific plans.

It might be true that “Moral is in the eye of the beholder”; everyone leads a daily life based on their moral standards. No exceptions at our domestic office. I had witnessed scads of irrational behaviors on a business trip back and forth and during my stay abroad.

――An employee who doesn’t care about putting the seal on slantly
――A management who lacks a marketing mentality
――An executive who misunderstands differentiation as instability of a bureaucratic organization

Too numerous to comprehensively list here.

I don’t know who their grandfather was; I am much more concerned with what their grandson will be.

Apart from joking, I spent my precious time talking those blinkards into changing their attitudes and minds many times over the last decade; however, nothing productive has happened as they don’t want to get out of such a hopeless lukewarm situation.

Sir Winston Churchill is quoted as saying, “Never, never, never, never give up.” With all due respect, my time outweighs their bland and blank time, so it’s not worth spending more time convincing them of their irrationality.

That being said, a good idea that proves someone is irrational has come to my mind. Believe it or not; however, it is not my intended meaning to be offensive to someone, so today, I challenge to prove an irrational expression by contradiction as below.

Prove by contradiction the expression \(\sqrt{3} + \sqrt{5}\) is an irrational number, providing \(\sqrt{3}\) and \(\sqrt{5}\) are respectively irrational.(Rref. Yellow Chart Math Ⅰ＋Ａ P74, slightly changed the formula)

Here is my solution to the problem.

Presume the expression \(\sqrt{3} + \sqrt{5}\) is not irrational, and it becomes a rational number.

Put \(\sqrt{3} + \sqrt{5} = a\) (\(a\) is a rational number), and \(\sqrt{5} = a – \sqrt{3}\). Square both sides, and \(5 = a^{2} – 2\sqrt{3}a + 3\). Then, \(2\sqrt{3}a = a^{2} – 2\).

Since \(a\neq 0\), \(\sqrt{3} = \frac{a^{2} – 2}{2a}\)・・・(＊)

Since \(a^{2} – 2\) and \(2a\) are rational, the right side of (＊) must be rational, which is contradictory to the fact that \(\sqrt{3}\) is irrational.

Therefore, the expression \(\sqrt{3} + \sqrt{5}\) is irrational.

Was the solution above a convincing rationale?

Stay tuned, and expect to see my next post.

Keep well.

Frank Yoshida

￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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