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It's Confusing.....

We know that,

\(\frac{1}{-1}\) = \(\frac{-1}{1}\)

Now taking square roots both sides,

\(\sqrt{\frac{1}{-1}}\) = \(\sqrt{\frac{-1}{1}}\)

Thus,

\(\frac{\sqrt{1}}{\sqrt{-1}}\) = \(\frac{\sqrt{-1}}{\sqrt{1}}\)

Multiplying \(\sqrt{1} \times \sqrt{-1}\) both sides,

\(\sqrt{1} \times \sqrt{1}\) = \(\sqrt{-1} \times \sqrt{-1}\)

Hence,

1 = -1

Can you explain this?????

Note by Sharad Roy
2 years, 8 months ago

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\(The\quad fallacy\quad occurs\quad in\quad going\quad from\quad \\ \sqrt { \frac { 1 }{ -1 } } to\quad \frac { \sqrt { 1 } }{ \sqrt { -1 } } \\ since\quad the\quad former\quad is\quad i\quad while\quad the\quad latter\\ is\quad -i.\quad \quad It's\quad not\quad always\quad true\quad that\\ \sqrt { \frac { a }{ b } } =\frac { \sqrt { a } }{ \sqrt { b } } \\ We\quad have\quad to\quad be\quad careful\quad when\quad doing\\ this,\quad when\quad signs\quad are\quad involved.\quad \) Michael Mendrin · 2 years, 8 months ago

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@Michael Mendrin It's always false when complex numbers are involved. Samuraiwarm Tsunayoshi · 2 years, 8 months ago

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