Please explain to me why this simple problem is so

[chambershex]

Disproving is much easier than proving. You only need one counter example to disprove that
"for all a and b a certain equality is satisfied". If you pick one counter example (say a=3 b=3), then the statement "for all..." is already violated.


I don't think there is a useful way to write
\sqrt{a+b}=\sqrt{a}+\sqrt{b}+x
But counterexamples like the ones above are really OK.

Hmmm... so there's no way to actually find the difference between the left and right sides of the following equation in terms of the variables a,b?

√(a + b) =/= √a + √b

It seems you can only find a difference by first getting rid of the root signs by squaring both sides (like Mentallic instructed me to do above). The difference found was 2√(ab), but that only refers to the difference between the modified equation!

You mention disproving is much easier than proving, but what I'm really having difficulty in understanding is the fact (as you said yourself) that "everything you learned can be derived from these [basic arithmetic] rules". If the whole of maths operates on a finite number of fundamental arithmetic rules (or axioms), then surely it follows that any expression, just like
√(a + b) =/= √a + √b, should be able to be directly proved using these principles. Rather than having to disprove by using methods such as simply substituting in numbers.

 

[chambershex]

Sure there is, x=\sqrt{a+b}-\sqrt{a}-\sqrt{b}

haha, unfortunately that doesn't shine much light on the situation ;)

 

[Mentallic]

haha, unfortunately that doesn't shine much light on the situation ;)

Agreed

Well what we have here is that \sqrt{a+b}<\sqrt{a}+\sqrt{b} for all a,b>0

Because from squaring we have, a+b<a+b+2\sqrt{ab}, 0<2\sqrt{ab}

Anyway, if you're looking for the difference between \sqrt{a+b} and \sqrt{a}+\sqrt{b} it's just going to be some modified version of what I gave.
or unless you can make something of \sqrt{a+b} and \sqrt{a+b+2\sqrt{ab}}

It would be tempting to say that since the difference between the modified (squared) expression is 2\sqrt{ab} then the difference between the unmodified expression is \sqrt{2\sqrt{ab}}=\sqrt[4]{4ab} but... it's not... so uh, yeah...

 

[chambershex]

It would be tempting to say that since the difference between the modified (squared) expression is 2\sqrt{ab} then the difference between the unmodified expression is \sqrt{2\sqrt{ab}}=\sqrt[4]{4ab} but... it's not... so uh, yeah...

Yeh I intially thought that but realized that was wrong. "It's just going to be some modified version of what I gave" - I thought that too, but I have absolutely no idea how one would derive the answer from that. I suspect it would get quite complicated (but I could easily be wrong), but whatever the case, I think somewhere in how the answer is derived would really shed light on the whole inital question of why the LHS and RHS expressions are different.

I find maths fascinating, but unfortunately my brain is far too feeble to truly understand its fundamental nature...

 

[Mentallic]

Yep, all you have to do is show \sqrt{a+b}-\sqrt{a}-\sqrt{b}\neq \sqrt[4]{4ab}

Um... isn't the difference of 2\sqrt{ab} from the squared expression enough to prove they're different? I think it is.

 

[chambershex]

Yes of course its enough to prove their different, but it doesn't explain WHY in the same sense that the original question I asked was described (with reference to the distributive law etc.). Also I wonder if its possible to represent the question geometrically much like the first one? But I don't want anyone to go to any more trouble, the main point of this thread was to just clarify some things for me conceptually, which it has done, and I thank everyone for their time.

 

[Tobias Funke]

Also I wonder if its possible to represent the question geometrically much like the first one? But I don't want anyone to go to any more trouble...

It's no trouble at all. Use Dickfore's drawing in post 17. Change x^2 to a and y^2 to b. Then the side of the entire square is sqrt(a)+sqrt(b).

Now the square of area a+b must be smaller than the entire square from the picture because the picture has those two extra rectangles. Smaller square means smaller side length, so sqrt(a+b)<=sqrt(a)+sqrt(b).

The reasoning is almost exactly the same.

 

[Mentallic]

Yes of course its enough to prove their different, but it doesn't explain WHY in the same sense that the original question I asked was described (with reference to the distributive law etc.). Also I wonder if its possible to represent the question geometrically much like the first one? But I don't want anyone to go to any more trouble, the main point of this thread was to just clarify some things for me conceptually, which it has done, and I thank everyone for their time.

We don't need to keep proving everything from scratch every single time. We build on what we know and use those facts to prove further things.

 

[Dickfore]

I just want to state on the record that this thread is bad and the OP should feel bad for starting it. In order to prove that something is not true, it is sufficient to find one particular counter-example. This was done in the first reply. Everything after this is nonsense.

 

[Tobias Funke]

I just want to state on the record that this thread is bad and the OP should feel bad for starting it.

Are you a Futurama fan by any chance?

"Your music is bad, and you should feel bad!"- Dr. Zoidberg

 

[Werg22]

Thanks for the response, but that's exactly the type of answer I didn't want (sorry if I wasn't clear). What you said shows both sides aren't equal but it doesn't tell me anything about WHY they aren't equal.

What do you mean why? In math we're interested in establishing true and false statements. A counter-example that disproves (x + y)^2 = x^2 + y^2 for some x and y shows that this statement is false. It ends there.

 

[The Chaz]

You guys are great teachers. How inspirational!

 

[chambershex]

I just want to state on the record that this thread is bad and the OP should feel bad for starting it. In order to prove that something is not true, it is sufficient to find one particular counter-example. This was done in the first reply. Everything after this is nonsense.

Haha I should feel bad for starting it? What is this a prestigious science journal? You should should feel bad for showing such pretentiousness. No one had to respond to this thread, but they chose to (including yourself). I already said before I got everything I wanted out this threa, so as far as I'm concerned it can be closed.

 

[Dickfore]

Haha I should feel bad for starting it? What is this a prestigious science journal? You should should feel bad for showing such pretentiousness. No one had to respond to this thread, but they chose to (including yourself). I already said before I got everything I wanted out this threa, so as far as I'm concerned it can be closed.

Yes, you managed to troll us.

 

[Hurkyl]

Honestly, the OP confounds me. The two sides appear different -- there should be no surprise to find out they actually are different.




Incidentally, the original inequality is not an identity -- it fails when x is 0 or when y is 0.

Inequalities, incidentally, cannot be proven by manipulating equalities. One must invoke something that is intrinsically non-algebraic¹, such as the inequality 0 \neq 1 of ordinary arithmetic.



1: I mean "algebraic" in a specific technical sense -- e.g. referring to universal algebra, or to essentially algebraic theories.

 

[alice22]

Well why don't you try it and find out the expansion

Well I had a go!

I can see the expansion pretty easily from the diagram although the diagram is not very good it does help visualise it.
You can see the 3x^2y 'slabs' that go on 3 sides of x^2 and the 3 y^2x 'columns which go on 3 vertices

.

 

[Dickfore]

Well I had a go!

I can see the expansion pretty easily from the diagram although the diagram is not very good it does help visualise it.

Could you please draw a diagram like this for the expansion of (x + y)^{4}. Thanks. Appreciate it.

 

[chambershex]

Yes, you managed to troll us.

Does the 'dick' in 'dickfore' happen to be a self-description by any chance? And btw finding a question extremely easy does not mean I boast the same intellect as you. So please, calling something 'trolling' because you simply find it trivial is really the height of arrogance.

EDIT: I am also aware of the irony that I am now actually trolling so don't bother to point that out.

 

[chambershex]

Inequalities, incidentally, cannot be proven by manipulating equalities. One must invoke something that is intrinsically non-algebraic¹, such as the inequality 0 \neq 1 of ordinary arithmetic.



1: I mean "algebraic" in a specific technical sense -- e.g. referring to universal algebra, or to essentially algebraic theories.

That sounds very interesting, and no doubt very complicated ;)

 

[Dickfore]

chambershex, how come x + y \ne x \, y? I know that, for example, 2 + 3 = 5 and 2 \cdot 3 = 6 and 5 \ne 6, but what's the reason behind this?

 

[Dickfore]

Unfortunately I cannot find a 4D drawing package!

Your previous diagram was not three dimensional either.

 

[The Chaz]

chambershex, how come x + y \ne x \, y? I know that, for example, 2 + 3 = 5 and 2 \cdot 3 = 6 and 5 \ne 6, but what's the reason behind this?

Why are you being so insulting? You and H must be on a mission to demoralize and deter people from using PF. Sure, the OP doesn't think like us or know as much, but who here has explained WHY?!? I suspect the reason that no one has (explained why) is that you don't know. The first response was close... Something is not true unless it is ALWAYS true.

As for your question, they are different operations, and have different outputs.

 

[alice22]

Your previous diagram was not three dimensional either.

No it was a 2 dimensional representation of the vertices of a 3D shape to be precise, or at least my rather poor attempt at it.

There is a video on the web claiming to draw a representation of a 4D cube but I think I will be banned if I post a link to it so I won't, as I expect it may break forum rules, or something like that!

 

[Dickfore]

Why are you being so insulting? You and H must be on a mission to demoralize and deter people from using PF. Sure, the OP doesn't think like us or know as much, but who here has explained WHY?!? I suspect the reason that no one has (explained why) is that you don't know. The first response was close... Something is not true unless it is ALWAYS true.

As for your question, they are different operations, and have different outputs.

But, for example, when x = y = 2 or x = y = 0 I get the same answer.

 

[chambershex]

chambershex, how come x + y \ne x \, y? I know that, for example, 2 + 3 = 5 and 2 \cdot 3 = 6 and 5 \ne 6, but what's the reason behind this?

I know your being nasty here, but I can answer that. The operation + is fundamental and cannot be broken down into anything more simple - what it means is self-evident. The operation x is derived from +, being defined as Y lots of X. Or alternatively X + X + X..., Y times. However, I'm sure someone could provide a better formal description.

 

[Dickfore]

I know your being nasty here, but I can answer that. The operation + is fundamental and cannot be broken down into anything more simple - what it means is self-evident. The operation x is derived from +, being defined as Y lots of X. Or alternatively X + X + X..., Y times. However, I'm sure someone could provide a better formal description.

Oooh! Sounds too complicated ;) What does operation mean?

 

[chambershex]

Why are you being so insulting? You and H must be on a mission to demoralize and deter people from using PF. Sure, the OP doesn't think like us or know as much, but who here has explained WHY?!? I suspect the reason that no one has (explained why) is that you don't know. The first response was close... Something is not true unless it is ALWAYS true.

As for your question, they are different operations, and have different outputs.

Thank-you. I now understand why the first question I posed is true, however I'm still yet to really see a good, fundamental explanation of why the second question is true (the one referring to the roots).

The overarching reason I am asking these questions is because I have always had a problem with completing algebraic/arithmetical questions using explanations which are just accepted as truth. I am content enough once an explanation can get down to the level of the axiom, because I can accept certain fundamental, propositions are self-evident. "I suspect the reason that no one has (explained why) is that you don't know" - I think this statement is actually not far from the truth; on philosophical grounds much of the foundations of mathematics is problematic.

 

[The Chaz]

But, for example, when x = y = 2 or x = y = 0 I get the same answer.

No, 4 is not the same as 0. Surprised you didn't know that.
/trollfeeding

 

[Dickfore]

Thank-you. I now understand why the first question I posed is true, however I'm still yet to really see a good, fundamental explanation of why the second question is true (the one referring to the roots).

The overarching reason I am asking these questions is because I have always had a problem with completing algebraic/arithmetical questions using explanations which are just accepted as truth. I am content enough once an explanation can get down to the level of the axiom, because I can accept certain fundamental, propositions are self-evident. "I suspect the reason that no one has (explained why) is that you don't know" - I think this statement is actually not far from the truth; on philosophical grounds much of the foundations of mathematics is problematic.

Let us say that the second statement you were asking about was correct, i.e.

<br /> \sqrt{a + b} = \sqrt{a} + \sqrt{b}<br />

was correct. Then, take a = x^{2} and b = y^{2} for x, y &gt; 0. Our 'equality' becomes:

<br /> \sqrt{x^{2} + y^{2}} = x + y<br />

Then, square the above 'equality':

<br /> x^{2} + y^{2} = (x + y)^{2}<br />

But, you this is in contradiction with your first statement. Therefore, if you accept that the last equality does not hold, then neither can the second one.