What is wrong?

Algebra Level 2

Here is a "proof" that 1 = 1 -1 = 1 :

  • 1. 1 = 1 2 1 = \sqrt { { 1 }^{ 2 } }

We know that:

  • 2. 1 2 = ( 1 ) 2 { 1 }^{ 2 } = { \left( -1 \right) }^{ 2 }

Substituting the 1. through the 2. leads to:

  • 3. 1 2 = ( 1 ) 2 \sqrt { { 1 }^{ 2 } } =\sqrt { { \left( -1 \right) }^{ 2 } }

Due to the power law a n m = a n m \sqrt [ m ]{ { a }^{ n } } = { a }^{ \frac { n }{ m } } we get:

  • 4. ( 1 ) 2 = ( 1 ) 2 2 \sqrt { { \left( -1 \right) }^{ 2 } } ={ \left( -1 \right) }^{ \frac { 2 }{ 2 } }

And finally simplifying the result.

  • 5. ( 1 ) 2 2 = ( 1 ) 1 = 1 { \left( -1 \right) }^{ \frac { 2 }{ 2 } }{ ={ \left( -1 \right) }^{ 1 }=-1 }

    Thus 1 = 1 q . e . d -1 = 1 \quad q.e.d

Which step is the first one, which is wrong?

3 2 1 4 5

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

CodeCrafter 1
May 29, 2019

The power law a n m = a n m \sqrt [ m ]{ { a }^{ n } } ={ a }^{ \frac { n }{ m } } is only defined for non negative real numbers. Thus step 4 is wrong.

Also mention that a 2 = a \sqrt { { a }^{ 2 } } =\left| a \right| And the absolute value of 1 -1 is 1 1 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...