【数学I】例題1.3.1：不等式の性質（One More）★★

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% 例題I1.3.1：不等式の性質（One More）★★
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$-2<x<1,1<y<3$のとき，次の式のとりうる値の範囲を求めよ．

(1) $x+1$

(2) $2x$

(3) $x+y$

(4) $x-y$

(5) $2x-3y$

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% 例題I1.3.1の解答（One More）★★
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(1) $-2<x<1$の各辺に$1$を加えると，$-1<x+1<2$

(2) $-2<x<1$の各辺に$2$を掛けると，$-4<2x<2$

(3) $-2<x<1$の各辺に$y$を加えると，$-2+y<x+y<1+y$

$1<y$より，$-2+1<-2+y$

また，$y<3$より，$1+y<1+3$

したがって，$-1<x+y,x+y<4$

よって，$-1<x+y<4$

(4) $1<y<3$の各辺に$-1$を掛けると，$-1>-y>-3$

すなわち，$-3<-y<-1$

したがって，$-2<x<1,-3<-y<-1$より，
\[
-2+(-3)<x+(-y)<1+(-1)
\]

よって，$-5<x-y<0$

(5) (2)より，$-4<2x<2$

$1<y<3$の各辺に$-3$を掛けると，$-3>-3y>-9$

すなわち，$-9<-3y<-3$

したがって，$-4<2x<2,-9<-3y<-3$より，
\[
-4+(-9)<2x+(-3y)<2+(-3)
\]

よって，$-13<2x-3y<-1$

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% 問題I1.3.1：不等式の性質（One More）★★
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$-3<x<2,-1<y<4$のとき，次の式のとりうる値の範囲を求めよ．

(1) $x+2$

(2) $3x$

(3) $x+y$

(4) $x-y$

(5) $3x-2y$

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% 問題I1.3.1の解答（One More）★★
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(1) $-3<x<2$の各辺に$2$を加えると，$-1<x+2<4$

(2) $-3<x<2$の各辺に$3$を掛けると，$-9<3x<6$

(3) $-3<x<2$の各辺に$y$を加えると，$-3+y<x+y<2+y$

$-1<y$より，$-3+(-1)<-3+y$

また，$y<4$より，$2+y<2+4$

したがって，$-4<x+y,x+y<6$

よって，$-4<x+y<6$

(4) $-1<y<4$の各辺に$-1$を掛けると，$1>-y>-4$

すなわち，$-4<-y<1$

したがって，$-3<x<2,-4<-y<1$より，$-3+(-4)<x+(-y)<2+1$

よって，$-7<x-y<3$

(5) (2)より，$-9<3x<6$

$-1<y<4$の各辺に$-2$を掛けると，$2>-2y>-8$

すなわち，$-8<-2y<2$

したがって，$-9<3x<6,-8<-2y<2$より，$-9+(-8)<3x+(-2y)<6+2$よって，$-17<3x-2y<8$

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% 例題I1.3.1の側注：不等式の性質（One More）★★
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% 側注1
$a<b\Rightarrow a+c<b+c$

% 側注2
$c>0,a<b\Rightarrow ac<bc$

% 側注3
$-2<x<1,1<y<3$の各辺を足し合わせて，$-1<x+y<4$としてもよい．

% 側注4
不等式の両辺に負の数を掛けるときは，不等号の向きが変わる．

% 側注5
$-2-1<x-y<1-3$より，$-3<x-y<-2$とする操作は誤りであるので注意すること．

% 側注6
不等号の向きが変わる．

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% 問題I1.3.1の側注：不等式の性質（One More）★★
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% 側注1
$a<b\Rightarrow a+c<b+c$

% 側注2
$c>0,a<b\Rightarrow ac<bc$

% 側注3
$-3<x<2,-1<y<4$の各辺を足し合わせて，$-4<x+y<6$としてもよい．

% 側注4
不等式の両辺に負の数を掛けるときは，不等号の向きが変わる．

% 側注5
不等号の向きが変わる．

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% 例題I1.3.1の考え方・補足など：不等式の性質（One More）★★
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% 考え方
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(1)は，不等式の性質により，$a<b\Longrightarrow a+c<b+c$であるから，$\{\begin{aligned}-2+1<&x+1\\&x+1<1+1\end{aligned}$

つまり，$-2<x<1$の各辺に$1$を加えた，$-1<x+1<2$が成り立つことを用いる．

(3)，(4)，(5)は$a<x<b,c<y<d\Longrightarrow a+c<x+y<b+d$などの不等式の性質を用いる．

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$a<b,c<d\Longrightarrow a+c<b+d,$

$a<b,c<d\Longrightarrow a-d<b-c,$

$0<a<b,0<c<d\Longrightarrow ac<bd$

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% 注意
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$a<x<b,c<y<d$のとき，$a-c<x-y<b-d$は必ずしも成り立たない．例えば，$-2<x<1,1<y<5$とすると，$-3<x-y<-4$となりこれは明らかに成り立たない．(4)のように$x+(-y)$として，$a-d<x-y<b-c$とする操作が正しいので注意すること．