# 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 $$\displaystyle \frac{1}{a+b}$$?