グローバルビジネスで役立つ数学（68）整数全体の集合の整合性（英語版）

Hi there!

I’m wondering why we, the Japanese people, are so sensitive to what other countries people talk about us. Do we always want to be nice guys to others to avoid any conflicts between the parties concerned?

We shouldn’t be too much afraid of being disliked by others. I neither trust anyone who tries to please everyone nor any country that takes please-everyone political policies. What’s wrong with a remark that the Japanese people are inscrutable?

Let’s enjoy debating political and diplomatic issues with other non-Japanese people and be assertive without shying away from arguments that might occur over us.

It is safe to say that the Japanese are comparatively thoughtful and considerate to others and put a lot of weight on tacit understanding. In the global community, however, silence is not always golden. Let’s talk candidly about why we’re happy or unhappy with our counterparts; many things cannot be reasoned in world affairs or our daily lives, and that’s why we need to talk face to face.

You might be ambivalent about what I’m driving at; don’t worry. Today, let’s challenge the integer word problem with its divisible numbers of a universal set in mathematics, where we reach a clear answer.

We have two universal sets of integers: Class \(A\) containing integers divisible by 15, and the other Class \(B\) with those divisible by 25. Also \(C =\) {\(x + y|x\in A, y\in B\)}. Prove the Class \(C\) conforms to a universal set including integers divisible by 5. (Source: Yellow Chart Math Ⅰ＋Ａ similar question made from P68)

Here is my solution to the problem.

Let’s name a universal set of integers divisible by 5, D.

(1) Firstly, we prove \(C\subset D\).
Regarding a discretional \(z\in C\), we have \(z = x + y, x = 15l, y = 25m\) (\(l,\) and \(m\) are integers).

\(z = 15l + 25m = 5(3l + 5m)\).
Since \(3l + 5m\) is an integer, so \(z\) becomes divisible by 5, so \(z\in D\).
Therefore, \(C\subset D\)・・・・・(Ⅰ)

(2) Secondly, we prove \(C\supset D\).
Regarding a discretion \(z\in D\), we have \(z = 5n\) (\(n\) is an integer), then \(5n = 30m + (-25n)\).

Since \(30m\in A, -25n\in B\), we put \(x = 30n, y = -25n\), so \(z = x + y\), then \(z\in C\).
Therefore, \(C\supset D\)・・・・・(Ⅱ)

Finally, we reach \(C = D\) from (Ⅰ) and (Ⅱ), which means the Class \(C\) conforms to a universal set including integers divisible by 5.

Life is too short for perseverance. Just fire away what you’re feeling.

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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