The problem is to find $\sqrt{(-1)\cdot(-1)}$
Approach 1 - $\sqrt{(-1)\cdot (-1)} = \sqrt{(-1)^2} = -1$
Approach 2 - $\sqrt{(-1)\cdot (-1)} = \sqrt{1} = 1$
Which is correct and why?
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Ludolila
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Prabhdeep Singh
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This already has an answer on the site. – Karl Feb 05 '17 at 09:11
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I can't see it, mind linking it? – Prabhdeep Singh Feb 05 '17 at 09:12
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Related: https://math.stackexchange.com/q/461695/115115, and from there http://math.stackexchange.com/questions/438/why-sqrt-1-times-1-neq-sqrt-12, http://math.stackexchange.com/questions/49169/why-sqrt-1-times-1-neq-sqrt-12 – Lutz Lehmann Feb 05 '17 at 12:02
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Approach 1 is the correct answer. Hint: 1 x 1 is not a square. – LAAE Feb 05 '17 at 12:39
2 Answers
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When dealing with the imaginary unit, you need to be careful using properties such as:
$$ \sqrt{ a\cdot b } = \sqrt{a} \cdot \sqrt{b} $$
This holds for all $ a,b \in \mathbb{R}_+ \cup \{0\} $, but not for negative real numbers. Therefore, the wrong is the first:
$$ \sqrt{ (-1)\cdot (-1)} \neq i \cdot i $$
Berationalgetreal
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I can understand this because we could get a negative answer in this way for every square root, but what is the fundamental mistake? Saying you can't do that is not really helpful – Prabhdeep Singh Feb 05 '17 at 08:57
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$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$
where a ≥ 0, b ≥ 0 Or a ≥ 0, b < 0
But NOT a < 0, b < 0
So applying in on a = b = -1 is invalid.
Kanwaljit Singh
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