is square root of 1 equal 1?

Aero763

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

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.

theodds

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 $$\displaystyle \sqrt{y}$$.

Aero763

Aero763

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.

TheCoffeeMachine

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)

Bacterius

Wilmer

2) And if so, then $$\displaystyle 1 + \sqrt{x} = \sqrt{1} + \sqrt{x} = \sqrt{1 + x}$$ ?
Is 1 + sqrt(4) = sqrt(5)?

undefined

MHF Hall of Honor
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 $$\displaystyle \frac{1}{a+b}$$?