The AF Problem

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If something over 0 is undefined, and if a part of an equation satisfies this condition, and if x=2 is just equal to 1/x=1/2 by cross-multiplying, then (x ^ { 3 } + y ^ { 3 } ) = 0 is equal to 1 / ( x ^ { 3 } + y ^ { 3 } ) = 1/0 by cross-multiplying giving and equation of 1 / ( x ^ { 3 } + y ^ { 3 } ) - 1/0 = 0; after doing least common denominator the result would be [ 0 - ( x ^ { 3 } + y ^ { 3 })] / 0 =0 which is - ( x ^ { 3 } + y ^ { 3 } ) /0 = 0, but it should not be equal to zero but undefined? Though if you solve it properly there are many solutions for this. Does it mean that all equations equal to 0 is undefined at the same time having a solution set? If not please explain why.

I could explain it in the comments I think it is right after all. WHAT A DISCOVERY!!! I do not care I could not explain it

Adrian Sing by Adrian Sing , 23, Philippines

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1 solution

Dpk ­
Jul 23, 2014

I'll assume this is your question:

I f s o m e t h i n g o v e r 0 i s u n d e f i n e d , a n d i f a p a r t o f a n e q u a t i o n s a t i s f i e s t h i s c o n d i t i o n , a n d i f x = 2 i s j u s t e q u a l t o 1 x = 1 2 b y c r o s s − m u l t i p l y i n g , t h e n ( x 3 + y 3 ) = 0 i s e q u a l t o 1 x 3 + y 3 = 1 0 b y c r o s s − m u l t i p l y i n g g i v i n g a n e q u a t i o n o f 1 ( x 3 + y 3 ) − 1 0 = 0 ; a f t e r d o i n g l e a s t c o m m o n d e n o m i n a t o r t h e r e s u l t w o u l d b e 0 − ( x 3 + y 3 ) 0 = 0 w h i c h i s − ( x 3 + y 3 ) 0 = 0 , b u t i t s h o u l d n o t b e e q u a l t o z e r o b u t u n d e f i n e d ? T h o u g h i f y o u s o l v e i t p r o p e r l y t h e r e a r e m a n y s o l u t i o n s f o r t h i s . D o e s i t m e a n t h a t a l l e q u a t i o n s e q u a l t o 0 i s u n d e f i n e d a t t h e s a m e t i m e h a v i n g a s o l u t i o n s e t ? I f n o t p l e a s e e x p l a i n w h y . If\quad something\quad over\quad 0\quad is\quad undefined,\quad and\quad if\quad a\quad part\quad of\quad an\quad equation\quad satisfies\quad this\quad condition,\\ and\quad if\quad x\quad =\quad 2\quad is\quad just\quad equal\quad to\quad \frac { 1 }{ x } \quad =\quad \frac { 1 }{ 2 } \quad by\quad cross-multiplying,\quad \\ then\quad (x^{ { 3 } }+y^{ { 3 } })\quad =\quad 0\quad is\quad equal\quad to\quad \frac { 1 }{ x^{ { 3 } }+y^{ { 3 } } } =\quad \frac { 1 }{ 0 } \quad by\quad cross-multiplying\quad giving\quad an\quad equation\quad of\\ \frac { 1 }{ (x^{ { 3 } }+y^{ { 3 } }) } -\frac { 1 }{ 0 } =\quad 0;\quad after\quad doing\quad least\quad common\quad denominator\quad the\quad result\quad would\quad be\quad \\ \frac { 0-(x^{ { 3 } }+y^{ { 3 } }) }{ 0 } \quad =\quad 0\quad which\quad is\quad \frac { -(x^{ { 3 } }+y^{ { 3 } }) }{ 0 } =\quad 0,\quad but\quad it\quad should\quad not\quad be\quad equal\quad to\quad zero\quad but\quad undefined?\\ Though\quad if\quad you\quad solve\quad it\quad properly\quad there\quad are\quad many\quad solutions\quad for\quad this.\\ Does\quad it\quad mean\quad that\quad all\quad equations\quad equal\quad to\quad 0\quad is\quad undefined\quad at\quad the\quad same\quad time\quad having\quad a\quad solution\quad set?\\ If\quad not\quad please\quad explain\quad why.

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and the answer:

there's a theorem (forgot what exactly what it was) that says the Solution Set of an expression is a subset of the solution set of the expression raised to a higher power.

Example:

x = 5 ; S S : { 5 } x\quad =\quad 5;\quad SS:\quad \left\{ 5 \right\}

squaring both sides results to:

x 2 = 25 ; S S : { − 5 , 5 } { x }^{ 2 }\quad =\quad 25;\quad SS:\quad \left\{ -5,5 \right\}

we can see that { 5 } \left\{ 5 \right\} is a subset of { − 5 , 5 } \left\{ -5,5 \right\}

What this means is that if you go up a power, the solution set expands, and if you go down, the solution set shrinks.

When you inverted both sides, you are essentially raising it to a negative power, and because of that, the solution set shrinks (now there's no more solution)

Besides, you're not allowed to invert both sides since it would involve dividing by zero at some point, which is not allowed.

DPK ­ - 6 years, 10 months ago

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