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

f o r s o l v i n g e q u a t i o n w h i c h o n e w e u s e

\Rightarrow a n d =, i m e a n w h e r e w e u s e \Rightarrow a n d w h e r e w e u s e, =

a n d w h e r e w e u s e o n e o f t h e m t o

c o n s i d e r w r o n g .

Answered by TheHoneyCat last updated on 21/Oct/21

When you write a sentence S_{1} that can be

either true or false,

f o r e x a m p l e “ I t i s {r a i n i n g}^{″} o r “ I a m {w e t}^{″}

o r “ x i s a s o l u t i o n t o t h e {e q u a t i o n}^{″} o r “ t h e

f u n c t i o n f i s {l i n e a r}^{″}

It is usulay assumed you belive that S_{1} is true

So if I write

x = {b l a b l a}_{1}

= {b l a b l a}_{2}

= {b l a b l a}_{3}

I, in fact, mean that

“ x = {b l a b l a}_{1}^{″}

and that “ x = {b l a b l a}_{2}^{″}

and that “ x = {b l a b l a}_{3}^{″}

This is the most common use of the sign “ =^{″}

“ \Leftrightarrow^{″} is very different, it is not simply used in

sentences, like “ =^{″}, it is used to make

sentences about other sentences .

“ S_{1} \Leftrightarrow 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

i f x^{2} = 2

y o u {d o n}^{'} t k n o w i f “ x = - {\sqrt{2}}^{″} o r n o t

i t c o u l d b e t r u e, b e c a u s e “ {(- \sqrt{2})}^{2} = 2^{″} i s t r u e

b u t x c o u l d a l s o b e + \sqrt{2}

s o y o u m i g t h w r i t e s o m e t i n g a s

“ x ⩽ 0 \Leftrightarrow x = - {\sqrt{2}}^{″} a n d “ x ⩾ 0 \Leftrightarrow x = {\sqrt{2}}^{″}

N o w t h a t m i g t h s e e q u i t e u s e l e s s, b u t i t i s

v e r y u s e f u l l w h e n y o u w a n t t o s o l v e a

q u e s t i o n l i k e :

D o e s x = y ?

y o u m i g h t t h e n w r i t e :

x = y \Leftrightarrow x - y = 0

\Leftrightarrow {b l a b l a}_{1}

\Leftrightarrow {b l a b l a}_{2}

\Leftrightarrow {b l a b l a}_{3}

a n d t h e n y o u c h e c k i f b l a b l a 3 i s t r u e o r n o t

Commented by TheHoneyCat last updated on 21/Oct/21

H e r e i s a n e x a m p l e :

(m y a n s w e r e i n b l u e)

Find all the positive solutions x of the

following equation :

3 x^{3} + 4 x^{2} - 5 x = 2

3 x^{3} + 4 x^{2} - 5 x = 2

\Leftrightarrow 3 x^{3} + 4 x^{2} - 5 x - 2 = 0

But, I also know that :

(3 x + 1) (x - 1) (x + 2)

= (3 x + 1) (x^{2} + 2 x - x - 2)

= (3 x + 1) (x^{2} + x - 2)

= 3 x^{3} + 3 x^{2} - 6 x + x^{2} + x - 2

= 3 x^{3} + 4 x^{2} - 5 x - 2

So,

3 x^{3} + 4 x^{2} - 5 x = 2

\Leftrightarrow (3 x + 1) (x - 1) (x + 2) = 0

\Leftrightarrow 3 x + 1 = 0 o r x - 1 = 0 o r x + 2 = 0

because if a product is 0, one of its numbers is

too

3 x + 1 = 0 \Leftrightarrow x = \frac{- 1}{3}

x - 1 = 0 \Leftrightarrow x = 1

x + 2 = 0 \Leftrightarrow x = - 2

So :

3 x^{3} + 4 x^{2} - 5 x = 2 \Leftrightarrow x \in {1, - 2, - 1 / 3}

And since x must be pisitive, the solution

is : x = 1

See the difference ?

Commented by zakirullah last updated on 21/Oct/21

g r e a t

Answered by talminator2856791 last updated on 22/Oct/21

i think this is one interpretation :

a = b

same mathematical object

a \Rightarrow b

b \Rightarrow a

different mathematical objects

dependant on each other .