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

Note by Trishit Chandra 2 years, 1 month ago

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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. – Micah Wood · 2 years ago

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@Micah Wood – |n|=+-(sqrt(n^2))

I thnk you are partially right. – Trishit Chandra · 2 years ago

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## Comments

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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. – Micah Wood · 2 years ago

Log in to reply

I thnk you are partially right. – Trishit Chandra · 2 years ago

Log in to reply