Unique nonneg. square root for nonneg. real number

[nobahar]

Hello,
No exclamation mark after hello I'm afraid, I'm frustrated with the following proof.
Okay, here's the theorem first:
Every nonnegative real number a has a unique nonnegative square root.
Here's the start of the proof, pretty much as they have done it but slightly condensed, all equations are included: a=0 then $$\sqrt(a)$$=0. If a>0 let S be the set of all positive x such that $$x^2 \leq a$$. Since (1+a)²>a the number (1+a) is an upper bound for S. Also, S is non-empty because the number $$\frac{a}{(1+a)}$$ is in S. In fact $$a^2 \leq a(1+a)^2$$ and hence $$\frac{a^2}{(1+a)^2} \leq a$$. S has a least upper bound, b; $$b\geq \frac{a}{(1+a)}$$ so b>0. Therefore, the three possibilities are b²>a, b²<a or b²=a. It then goes on to prove b²=a after some further working out.
Okay, so $$x^2 \leq a$$ indicates the range that the elements x in set S can fall into. The next bit, (1+a)²>a, means (1+a) is the upper bound, a² can be equal to a and in that case the element x=a=1, so this is can't be an upper bound. So (1+a) forms the upper bound, but surely (a+tiny amount)² would be an upper bound as I move closer to a?
Next, S is not a void set because $$\frac{a}{(1+a)}$$ is in S. I figured since it has to be <a but >0, but it doesn't imediately seem an obvious 'move', if that makes sense. The next bit throws me, a²<=a(1+a)², and since I don't know what's going on here I can't proceed with the proof. It also may come across from what I have written, but I'm not confident with the first 'bits'. Anyone want to attempt to explain this for me? Not all the proof, preferably upto what I have included here if you don't mind.
Thanks.

 

[LightHero]

The key part of the proof is understanding why (1+a)2 > a. This is because when a is any nonnegative real number, then (1+a) must be greater than 1 and so (1+a)2 must be larger than a. The next part is showing that S is not empty. Since a is nonnegative, \frac{a}{(1+a)} is also nonnegative and is less than a, meaning it is an element of S. The last part of the proof relies on showing that a2 <= a(1+a)2. This is because a2 = a*a, and a(1+a)2 = a * (1+a) * (1+a). Since (1+a) is greater than 1, a(1+a)2 must be greater than or equal to a2. These steps together form the proof that every nonnegative real number has a unique nonnegative square root.

 

[Architeuthis Dux]

Hello,

I understand your frustration with the proof, as it can be confusing and overwhelming at first. Let me try to break it down for you.

First, the theorem states that every nonnegative real number has a unique nonnegative square root. This means that for any nonnegative real number a, there exists only one nonnegative number x such that x^2 = a.

Now, let's look at the proof. The first part deals with the case where a = 0. In this case, the square root of a is also 0. This is pretty straightforward.

Next, the proof looks at the case where a > 0. Here, we define a set S, which contains all the positive numbers x such that x^2 is less than or equal to a. This means that S includes numbers such as 1, 2, 3, etc. but also numbers like 0.5, 0.1, 0.01, etc. Basically, any positive number whose square is less than or equal to a.

The next statement, (1+a)^2 > a, is important because it tells us that the number (1+a) is an upper bound for S. This means that all the elements of S are less than or equal to (1+a). Now, you mentioned that (a+tiny amount)^2 would also be an upper bound. While this is true, it is not the least upper bound. The proof is trying to find the smallest possible upper bound for S, and (1+a) satisfies this condition.

Now, the proof states that S is non-empty because a/(1+a) is an element of S. This is because a/(1+a) is a positive number whose square is less than or equal to a. You are correct in saying that it is less than a but greater than 0. This may not seem like an obvious move, but it is a key step in the proof.

Next, the statement a^2 <= a(1+a)^2 is just expanding the expression (1+a)^2. This is important because it helps us understand the relationship between a and (1+a). And since we know that (1+a) is an upper bound for S, this inequality tells us that a^2 is also less than or equal to (1+a)^2.

The final part of the proof deals with the three possibilities for b^2 - a: b^2