# グローバルビジネスで役立つ数学（67）集合の要素を決定する（英語版）

Hi there!

As is often the case in the corporate world, internal politics plays a pivotal role as a sobering reality. I might have been mature or slightly perverse, and I was deeply into who had much clout in career development since I joined a trading company.

There were no obvious academic cliques; however, the management was divided into three factions from a strategical point of view. The top three leaders were President, Senior Executive Director, and Export Department, Director.

I was assigned to work directly under the Senior Executive Director several years after I had joined the company. One day, the director told me that he had picked me out from the members of the Export Department, for I was always talking back to my previous superiors.

The senior director was so harsh on me; he taught me in a draconian way how Shoushaman should be and how I can develop into an overseas sales representative of a high caliber.

Caution is advised. Even if you’re fresh out of college, you might want to have insightfulness, perseverance, and tenacity to make it happen. It cannot be challenging and rewarding for you as long as you work under an incompetent boss.

Today, instead of a corporate faction, let’s scrutinize how to determine the members of a Class in mathematics.

We have the two classes; $$A =$$ {$$2, 6, 5a – a^{2}$$} and $$B =$$ {$$3, 4, 3a – 1, a + b, 9$$} providing the members of each class should be integers. Also, $$A\cap B =$$ {$$4, 6$$}. (1) Evaluate the constant numbers $$a$$ and $$b$$. (2) Find the members of the inclusion relation $$A\cup B$$. (Source: Yellow Chart Math Ⅰ＋Ａ P66)

Here is my solution to the problem.

For the question numbered (1),
judging from $$A\cap B =$$ {$$4, 6$$}, we can understand $$4\in A$$, so
$$5a – a^{2} = 4$$, i.e., $$a^{2} – 5a + 4 = 0$$・・・・・(Ⅰ)

Also, from $$6\in B$$,
$$3a – 1 = 6$$・・・・・(Ⅱ), or
$$a + b = 6$$・・・・・(Ⅲ)

Factorize (Ⅰ) into $$(a – 1)(a – 4) = 0$$, and we get $$a = 1, 4$$.
However, this result does not satisfy (Ⅱ).

At $$a = 1$$, we get $$b = 5$$ from (Ⅲ). In this case, $$A =$$ {$$2, 6, 4$$}, $$B =$$ {$$3, 4, 2, 6$$}, which makes $$A\cap B =$$
{$$2, 4, 6$$} and does not satisfy the requirement given.

Whereas at $$a = 4$$, we get $$b = 2$$ from (Ⅲ). In this case, $$A =$$ {$$2, 6, 4$$}, $$B =$$ {$$3, 4, 11, 6$$}, which satisfies the requirement given.
Therefore, the answer is $$a = 4, b = 2$$.

For the question numbered (2),
we can quickly get the answer $$A\cup B =$$ {$$2, 3, 4, 6, 11$$}, given the answer (1): $$A =$$ {$$2, 6, 4$$} and $$B =$$
{$$3, 4, 11, 6$$}.

Now I have mellowed in old age, young at heart, though. I will catch the trade winds in my sails――Explore, Dream, Discover.

Stay tuned, and expect to see my next post.

Keep well.

Frank Yoshida

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【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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