is square root of 1 equal 1?

**Aero763** · June 10th 2010, 06:14 PM

1) Is $\sqrt{1} = 1$ ?

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

3) If 1 and 2 is true, then:

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

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

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

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

Why is that?

**Krizalid** · June 10th 2010, 06:19 PM

$\sqrt{a+b}=\sqrt a+\sqrt b$ is false, but it's true that $\sqrt{a+b}\le\sqrt a+\sqrt b.$

**Aero763** · June 10th 2010, 06:29 PM

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.

**theodds** · June 10th 2010, 06:33 PM

Originally Posted by Aero763

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 $\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ for arbitrary $a, b$ . In fact, from the last two lines you already have the inequality that Krizalid posted if you replace $1$ with $\sqrt{y}$ .

**Aero763** · June 10th 2010, 11:07 PM

Thank you guys.

I got which is the problem step was.

But what is the difference between $\sqrt{1} + \sqrt{x}$ and $\sqrt{1 + x}<br />$ ?

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.