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Question Number 194796 by sniper237 last updated on 15/Jul/23
A 1m^2  rectangle which length is   less than 1 is  a square. Why?
$${A}\:\mathrm{1}{m}^{\mathrm{2}} \:{rectangle}\:{which}\:{length}\:{is}\: \\ $$$${less}\:{than}\:\mathrm{1}\:{is}\:\:{a}\:{square}.\:{Why}? \\ $$
Answered by Frix last updated on 15/Jul/23
It′s obviously false  l×w=1  w=(1/l); 0<l<1 ⇒ w>1  ⇒ l≠w ⇒ it′s not a square
$$\mathrm{It}'\mathrm{s}\:\mathrm{obviously}\:\mathrm{false} \\ $$$${l}×{w}=\mathrm{1} \\ $$$${w}=\frac{\mathrm{1}}{{l}};\:\mathrm{0}<{l}<\mathrm{1}\:\Rightarrow\:{w}>\mathrm{1} \\ $$$$\Rightarrow\:{l}\neq{w}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\boldsymbol{{not}}\:\mathrm{a}\:\mathrm{square} \\ $$
Commented by sniper237 last updated on 15/Jul/23
0<w≤l≤1 .  So  w≤1 and (1/w) =l≤1   hence  1≤w≤1 , ie  w=1 and l=1
$$\mathrm{0}<{w}\leqslant{l}\leqslant\mathrm{1}\:.\:\:{So}\:\:{w}\leqslant\mathrm{1}\:{and}\:\frac{\mathrm{1}}{{w}}\:={l}\leqslant\mathrm{1}\: \\ $$$${hence}\:\:\mathrm{1}\leqslant{w}\leqslant\mathrm{1}\:,\:{ie}\:\:{w}=\mathrm{1}\:{and}\:{l}=\mathrm{1}\: \\ $$
Commented by Frix last updated on 15/Jul/23
But now l=1 and you stated “l is less than  1” ⇔ l<1
$$\mathrm{But}\:\mathrm{now}\:{l}=\mathrm{1}\:\mathrm{and}\:\mathrm{you}\:\mathrm{stated}\:“\mathrm{l}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than} \\ $$$$\mathrm{1}''\:\Leftrightarrow\:{l}<\mathrm{1} \\ $$

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