Hi there!
It’s been a long time since I was labeled “maverick.” You might ask me, “Don’t you feel lonely?” I haven’t wanted to be so, but my idiosyncrasy didn’t allow me to congregate with others.
I don’t know why, but I got along with the top cat of a student gang in my elementary and junior high school days. The group leader, particularly in junior high school, was groovy because he never tried to have a large following by saying:
――”Flocking in a group takes my coolness away.”
He reiterated the remark to his buddies.
My maverick spirit might have been fostered through my experiences as an expatriate or overseas sales representative. I still think a guy is groovy when he acts as a harbinger or stays aloof.
I’ve never been a follower of a group; however, today, grudgingly, let me look out to scrutinizing an inclusion relation in mathematics.
The whole real numbers are a universal set, given \(A =\){\(x|-1\leq x <5\)}
and \(C =\){\(x|k – 6 < x < k + 1\)} (\(k\) is a constant number). Find the range
of value \(k\) to make \(A\subset C\).
Here is my solution to the problem.
The conditions required to make \(A\subset C\) are:
\(k – 6 < -1\)・・・・・（Ⅰ）
\(5\leq k + 1\)・・・・・(Ⅱ)
The above (Ⅰ) and (Ⅱ) must stand, so
From (Ⅰ), \(k < 5\), and from (Ⅱ), \(4\leq k\).
Therefore, the range in common is \(4\leq k < 5\).
Solving the above inclusion relation problem hasn’t affected my disposition. I’m going to go on a solo motorcycle tour to keep aloof.
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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