is square root of 1 equal 1?

Jun 2010
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0
1) Is \(\displaystyle \sqrt{1} = 1\) ?



2) And if so, then \(\displaystyle 1 + \sqrt{x} = \sqrt{1} + \sqrt{x} = \sqrt{1 + x}\) ?

3) If 1 and 2 is true, then:


\(\displaystyle (1 + \sqrt{x})^2 = (\sqrt{1 + x})^2\)


But if you actually square \(\displaystyle (1 + \sqrt{x})^2 \) and \(\displaystyle (\sqrt{1 + x})^2\), you'll find out that they are different:

\(\displaystyle (1 + \sqrt{x})^2 = (1 + \sqrt{x})(1 + \sqrt{x}) = 1 + 2\sqrt{x} + x\)

\(\displaystyle (\sqrt{1 + x})^2 = 1 + x\)

Why is that?
 

Krizalid

MHF Hall of Honor
Mar 2007
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\(\displaystyle \sqrt{a+b}=\sqrt a+\sqrt b\) is false, but it's true that \(\displaystyle \sqrt{a+b}\le\sqrt a+\sqrt b.\)
 
Jun 2010
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That's intresting information. Maybe some links on the topic, where I can have some in-depth study of the situation (c) (all the rights to Jersey Shore) here? Youtube videos would be great, but some text explanation will do too.
 
Oct 2009
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That's intresting information. Maybe some links on the topic, where I can have some in-depth study of the situation (c) (all the rights to Jersey Shore) here? Youtube videos would be great, but some text explanation will do too.
Your original post is essentially a proof that \(\displaystyle \sqrt{a} + \sqrt{b} \ne \sqrt{a + b}\) for arbitrary \(\displaystyle a, b\). In fact, from the last two lines you already have the inequality that Krizalid posted if you replace \(\displaystyle 1\) with \(\displaystyle \sqrt{y}\).
 
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Jun 2010
13
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Thank you guys.

I got which is the problem step was.

But what is the difference between \(\displaystyle \sqrt{1} + \sqrt{x}\) and \(\displaystyle \sqrt{1 + x}
\) ?

I want to get some intuition: why it is different?

Because on the first glance they not that different: you can say they look basically the same.

So, let me get it straight:

What i got from this discussion - is that two numbers under one root - is not two numbers, but one.

Thank you, gentlemen.
 
Mar 2010
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381
But what is the difference between \(\displaystyle \sqrt{1} + \sqrt{x}\) and \(\displaystyle \sqrt{1 + x}
\)
I want to get some intuition: why it is different?
Plug \(\displaystyle x = 4: \;\;\;\;\;\;\;\;\; 1+\sqrt{4} = 1+2 = 3\) and \(\displaystyle \sqrt{1+4} = \sqrt{5}\). (Giggle)
 
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undefined

MHF Hall of Honor
Mar 2010
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Because on the first glance they not that different: you can say they look basically the same.
I don't understand this statement. I suppose \(\displaystyle \frac{1}{a}+\frac{1}{b}\) looks "basically the same" as \(\displaystyle \frac{1}{a+b}\)?
 
Jun 2010
19
0
Bali, Indonesia.
2) is wrong.
therefore, the next statements are wrong.