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Question Number 157272 by zakirullah last updated on 21/Oct/21

for solving equation which one we use ⇒ and =, i mean where we use ⇒ and where we use, = and where we use one of them to consider wrong.

$${for}\:{solving}\:{equation}\:{which}\:{one}\:{we}\:{use} \\ $$$$\Rightarrow\:{and}\:=,\:{i}\:{mean}\:{where}\:{we}\:{use}\:\Rightarrow\:{and}\:{where}\:{we}\:{use},\:= \\ $$$${and}\:{where}\:{we}\:{use}\:{one}\:{of}\:{them}\:{to} \\ $$$${consider}\:{wrong}. \\ $$

Answered by TheHoneyCat last updated on 21/Oct/21

When you write a sentence S_(1 ) that can be either true or false, for example “It is raining” or “I am wet” or “x is a solution to the equation” or “the function f is linear” It is usulay assumed you belive that S_1 is true So if I write x=blabla_1 =blabla_2 =blabla_3 I, in fact, mean that “x=blabla_1 ” and that “x=blabla_2 ” and that “x=blabla_3 ” This is the most common use of the sign “=” “⇔” is very different, it is not simply used in sentences, like “=”, it is used to make sentences about other sentences. “S_1 ⇔S_2 ” means that the sentence S_1 is true if S_2 is and wrong if S_2 is. So during a math proof (or any speech about some math), you may use it to this purpose. For example if x^2 =2 you don′t know if “x=−(√2)” or not it could be true, because “(−(√2))^2 =2” is true but x could also be +(√2) so you migth write someting as “x≤0⇔x=−(√2)” and “x≥0⇔x=(√2)” Now that migth see quite useless, but it is very usefull when you want to solve a question like: Does x=y? you might then write: x=y⇔x−y=0 ⇔blabla_1 ⇔blabla_2 ⇔blabla_3 and then you check if blabla3 is true or not

$$ \\ $$$$\mathrm{When}\:\mathrm{you}\:\mathrm{write}\:\mathrm{a}\:\mathrm{sentence}\:\mathrm{S}_{\mathrm{1}\:} \mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{either}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false},\: \\ $$$${for}\:{example}\:``{It}\:{is}\:{raining}''\:{or}\:``{I}\:{am}\:{wet}'' \\ $$$${or}\:``\mathrm{x}\:{is}\:{a}\:{solution}\:{to}\:{the}\:{equation}''\:{or}\:``{the} \\ $$$${function}\:\mathrm{f}\:{is}\:{linear}'' \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{usulay}\:\mathrm{assumed}\:\mathrm{you}\:\mathrm{belive}\:\mathrm{that}\:\mathrm{S}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{true} \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{if}\:\mathrm{I}\:\mathrm{write} \\ $$$${x}={blabla}_{\mathrm{1}} \\ $$$$={blabla}_{\mathrm{2}} \\ $$$$={blabla}_{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{I},\:\mathrm{in}\:\mathrm{fact},\:\mathrm{mean}\:\mathrm{that} \\ $$$$``{x}={blabla}_{\mathrm{1}} '' \\ $$$$\mathrm{and}\:\mathrm{that}\:``{x}={blabla}_{\mathrm{2}} '' \\ $$$$\mathrm{and}\:\mathrm{that}\:``{x}={blabla}_{\mathrm{3}} '' \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{is}\:\:\mathrm{the}\:\mathrm{most}\:\mathrm{common}\:\mathrm{use}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sign}\:``='' \\ $$$$ \\ $$$$ \\ $$$$``\Leftrightarrow''\:\mathrm{is}\:\mathrm{very}\:\mathrm{different},\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{simply}\:\mathrm{used}\:\mathrm{in} \\ $$$$\mathrm{sentences},\:\mathrm{like}\:``='',\:\mathrm{it}\:\mathrm{is}\:\mathrm{used}\:\mathrm{to}\:\mathrm{make} \\ $$$$\mathrm{sentences}\:\mathrm{about}\:\mathrm{other}\:\mathrm{sentences}. \\ $$$$``{S}_{\mathrm{1}} \Leftrightarrow{S}_{\mathrm{2}} ''\:\mathrm{means}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sentence}\:{S}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{if}\:{S}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{and}\:\mathrm{wrong}\:\mathrm{if}\:{S}_{\mathrm{2}} \:\mathrm{is}. \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{during}\:\mathrm{a}\:\mathrm{math}\:\mathrm{proof}\:\left(\mathrm{or}\:\mathrm{any}\:\mathrm{speech}\:\mathrm{about}\right. \\ $$$$\left.\mathrm{some}\:\mathrm{math}\right),\:\mathrm{you}\:\mathrm{may}\:\mathrm{use}\:\mathrm{it}\:\mathrm{to}\:\mathrm{this}\:\mathrm{purpose}. \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$${if}\:\:{x}^{\mathrm{2}} =\mathrm{2} \\ $$$${you}\:{don}'{t}\:{know}\:{if}\:``{x}=−\sqrt{\mathrm{2}}''\:{or}\:{not} \\ $$$${it}\:{could}\:{be}\:{true},\:{because}\:``\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}''\:{is}\:{true} \\ $$$${but}\:{x}\:{could}\:{also}\:{be}\:+\sqrt{\mathrm{2}} \\ $$$${so}\:{you}\:{migth}\:{write}\:{someting}\:{as} \\ $$$$``{x}\leqslant\mathrm{0}\Leftrightarrow{x}=−\sqrt{\mathrm{2}}''\:{and}\:``{x}\geqslant\mathrm{0}\Leftrightarrow{x}=\sqrt{\mathrm{2}}'' \\ $$$$ \\ $$$$ \\ $$$${Now}\:{that}\:{migth}\:{see}\:{quite}\:{useless},\:{but}\:{it}\:{is} \\ $$$${very}\:{usefull}\:{when}\:{you}\:{want}\:{to}\:{solve}\:{a} \\ $$$${question}\:{like}: \\ $$$${Does}\:{x}={y}? \\ $$$${you}\:{might}\:{then}\:{write}: \\ $$$${x}={y}\Leftrightarrow{x}−{y}=\mathrm{0} \\ $$$$\Leftrightarrow{blabla}_{\mathrm{1}} \\ $$$$\Leftrightarrow{blabla}_{\mathrm{2}} \\ $$$$\Leftrightarrow{blabla}_{\mathrm{3}} \\ $$$${and}\:{then}\:{you}\:{check}\:{if}\:{blabla}\mathrm{3}\:{is}\:{true}\:{or}\:{not} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TheHoneyCat last updated on 21/Oct/21

Here is an example: (my answere in blue) Find all the positive solutions x of the following equation: 3x^3 +4x^2 −5x=2 3x^3 +4x^2 −5x=2 ⇔3x^3 +4x^2 −5x−2=0 But, I also know that: (3x+1)(x−1)(x+2) =(3x+1)(x^2 +2x−x−2) =(3x+1)(x^2 +x−2) =3x^3 +3x^2 −6x+x^2 +x−2 =3x^3 +4x^2 −5x−2 So, 3x^3 +4x^2 −5x=2 ⇔(3x+1)(x−1)(x+2)=0 ⇔3x+1=0 or x−1=0 or x+2=0 because if a product is 0, one of its numbers is too 3x+1=0⇔x=((−1)/3) x−1=0⇔x=1 x+2=0⇔x=−2 So: 3x^3 +4x^2 −5x=2 ⇔ x∈{1,−2,−1/3} And since x must be pisitive, the solution is : x=1 See the difference?

$${Here}\:{is}\:{an}\:{example}: \\ $$$$\left({my}\:{answere}\:{in}\:{blue}\right) \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{solutions}\:{x}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{equation}: \\ $$$$\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{2}\: \\ $$$$ \\ $$$$\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{2} \\ $$$$\Leftrightarrow\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{But},\:\mathrm{I}\:\mathrm{also}\:\mathrm{know}\:\mathrm{that}: \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right) \\ $$$$=\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−{x}−\mathrm{2}\right) \\ $$$$=\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{2}\right) \\ $$$$=\mathrm{3}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+{x}^{\mathrm{2}} +{x}−\mathrm{2} \\ $$$$=\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2} \\ $$$$ \\ $$$$\mathrm{So}, \\ $$$$\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{2} \\ $$$$\Leftrightarrow\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\:{or}\:{x}−\mathrm{1}=\mathrm{0}\:{or}\:{x}+\mathrm{2}=\mathrm{0}\: \\ $$$$\mathrm{because}\:\mathrm{if}\:\mathrm{a}\:\mathrm{product}\:\mathrm{is}\:\mathrm{0},\:\mathrm{one}\:\mathrm{of}\:\mathrm{its}\:\mathrm{numbers}\:\mathrm{is} \\ $$$$\mathrm{too} \\ $$$$ \\ $$$$\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\Leftrightarrow{x}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${x}−\mathrm{1}=\mathrm{0}\Leftrightarrow{x}=\mathrm{1} \\ $$$${x}+\mathrm{2}=\mathrm{0}\Leftrightarrow{x}=−\mathrm{2} \\ $$$$ \\ $$$$\mathrm{So}: \\ $$$$\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{2}\:\Leftrightarrow\:{x}\in\left\{\mathrm{1},−\mathrm{2},−\mathrm{1}/\mathrm{3}\right\} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{since}\:{x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{pisitive},\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{is}\::\:{x}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{See}\:\mathrm{the}\:\mathrm{difference}? \\ $$

Commented by zakirullah last updated on 21/Oct/21

great

$${great} \\ $$

Answered by talminator2856791 last updated on 22/Oct/21

i think this is one interpretation: a = b same mathematical object a ⇒ b b ⇒ a different mathematical objects dependant on each other.

$$\:\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{one}\:\mathrm{interpretation}:\:\: \\ $$$$\: \\ $$$$\:\mathrm{a}\:=\:\mathrm{b} \\ $$$$\:\mathrm{same}\:\mathrm{mathematical}\:\mathrm{object}\:\: \\ $$$$\: \\ $$$$\:\mathrm{a}\:\Rightarrow\:\mathrm{b} \\ $$$$\:\mathrm{b}\:\Rightarrow\:\mathrm{a} \\ $$$$\:\mathrm{different}\:\mathrm{mathematical}\:\mathrm{objects}\:\:\:\: \\ $$$$\:\mathrm{dependant}\:\mathrm{on}\:\mathrm{each}\:\mathrm{other}. \\ $$

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