New user? Sign up

Existing user? Sign in

Is this correct...????????

Note by Trishit Chandra 2 years, 7 months ago

Sort by:

You just played around with fire.

You made some foul moves. One example is that you replaced \(1\) to \(i^4\)then took square root of it. Reason is this:

Clearly \(\sqrt{1} = 1\). Now since \(i^4 =1\), we have \(\sqrt{1} = \sqrt{i^4} = i^2 = -1\). So we have \(1=-1\).

We know \(1\) is not supposed to be equal to \(-1\).

Also \(\sqrt{n^2}\) is actually equal to \(|n|\), not \(\pm n\), since \(n^2\ge0\) for all real numbers.

Log in to reply

|n|=+-(sqrt(n^2))

I thnk you are partially right.

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestYou just played around with fire.

You made some foul moves. One example is that you replaced \(1\) to \(i^4\)then took square root of it. Reason is this:

Clearly \(\sqrt{1} = 1\). Now since \(i^4 =1\), we have \(\sqrt{1} = \sqrt{i^4} = i^2 = -1\). So we have \(1=-1\).

We know \(1\) is not supposed to be equal to \(-1\).

Also \(\sqrt{n^2}\) is actually equal to \(|n|\), not \(\pm n\), since \(n^2\ge0\) for all real numbers.

Log in to reply

|n|=+-(sqrt(n^2))

I thnk you are partially right.

Log in to reply