Please explain to me why this simple problem is so

[chambershex]

Hey,

I just wanted to know if someone could explain to me why the following is true:

(x+y)^2 =/= x^2 + y^2

So, in words, why doesn't the some of two numbers added together and then squared equal the sum of the numbers squared individually and then added together?

I obviously know how to show they are not equal by simply explanding the left side, but I'm looking for a really fundamental answer (i.e. is there some axiom which accounts for this?).

Thanks

 

[Mentallic]

Take any number, say x=2 and y=3, now plug into your equality. Obviously they aren't the same numbers so how could both left and right sides be equal to each other then?

 

[chambershex]

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.

 

[uart]

I'm still a bit confused about exactly what type of answer you want, but what about the explanation :

"Order of operations" is important in mathematics. That is, in general you cannot exchange the order of operations in a mathematical expression and always expect the result to be unchanged.

 

[chambershex]

I'm still a bit confused about exactly what type of answer you want, but what about the explanation :

"Order of operations" is important in mathematics. That is, in general you cannot exchange the order of operations in a mathematical expression and always expect the result to be unchanged.

That answer might be leading in the right direction...Maybe it will help if I frame the question with reference to the order of operations:

WHY does squaring two numbers after they have first been added together produce a result different to when they are squared individually first and then added together?

 

[Mentallic]

Sorry I thought you were just looking for some answer other than "expand the Left side and you'll see".

If the operation of expanding is clearly defined, then you shouldn't have any doubts.

$$a(x+y)=ax+ay$$, yes?

Well, let a=x+y, Now we have $$(x+y)(x+y)=(x+y)x+(x+y)y=x^2+2xy+y^2=(x+y)^2\neq x^2+y^2$$

Are you convinced? Else the only other way I can answer your question is with "just because".

 

[Mentallic]

or even take a geometrical approach to the problem:

Have some length x, and then add some length y to the end of it so now you have x+y length. Now square this (actually make it a square of x+y length and width). Notice how this square can be split up into 4 sections, x², y² and the last two are xy each. Obviously now (x+y)²>x²+y² since we're dealing with positive lengths x and y. Except of course in the obvious case that x=0 or y=0, $$(x+y)^2\neq x^2+y^2$$

 

[FieldDuck]

You can just view this as distribution.
(x+y)(x+y) = y(x+y) + x(x+y) = xy + y^2 + x^2 + xy = x^2 + 2xy + y^2 so if your looking for something "fundamental" as to why you would expand the left side as you do, then it's just really distribution law.

Edit: oops someone got to it before I did. Mentallic is correct.

 

[Mentallic]

Mentallic is correct.

I hope so

 

[Jerbearrrrrr]

All of maths "just is" because it is.
Maybe that's why it's hard to explain elementary arithmetic/algebra.

Or stuff like "why isn't (uv)' equal to u' v'?" where ' denotes d/dx. It just...isn't, because we know how to derive the correct answer and everything is a counterexample; and why should it be? Argh lol.

Sometimes you get asked things like "Why is -x^2 not (-x)^2" and the only reasons you can give are basically 'just apply stuff you know', or 'assume it is, then show x=0 or a contradiction'. However, no one (except someone who already gets it anyway) will listen to an answer like the second one.

 

[freireib]

Here is a good link with a nice geometric explanation.
http://www.tjleone.com/BinomialSquareExplained.pdf

 

[Gerenuk]

The answer is very simple: First tell me why you think you can do it your way!

Maths works like this: It is very well defined what is are allowed to do. It's about 20 rules of algebra you learned at school. And anything else is definitely wrong!
In fact any manipulation you ever did in school can be traced back to about 10 laws for basic arithmetics. The rules look like
http://en.wikipedia.org/wiki/Ring_(mathematics)#Formal_definition
Some of these you might find even too trivial, but I assure you that everything you learned can be derived from these rules (provided you define notation).

It's the single most serious mistake to make up new rules. See it this way:
If there are 10 real rules and you know 9 of then, then you will get 90% of the answers correct and 10% you won't find - at least you know which ones.

If however you know the 10 rules and one additional erroneous 11th rules, then you get all results wrong, because your wrong rule spoils everything like a virus.

So I'd rather want to know my student one rule too little, than one rule too much.

 

[Integral]

Here is a good link with a nice geometric explanation.
http://www.tjleone.com/BinomialSquareExplained.pdf

The first figure in this link should provide the answer you need. call the side length of the blue square a, while b is the side length of the red square. So the blue square has area a² the area of the red square is b², the larger square has side length of (a+b) and area of (a+b)². You should be able to see that the sum of a² + b² is not the same as (a+b)². Note that the difference is 2 rectangles with area ab. That gives the total area of the big square as a²+ b² + 2ab. This is the exact result of algebracally expanding (a+b)².

 

[chambershex]

Thanks everyone for the input. I was thinking along the lines of the distributive law, which Mentallic later verified, and I guess it does seem to be pretty much as fundamental as you can go. But, of course, the question of why the distributive law is so can then be raised...

Thanks freireib too, that was an excellent summary of all the posts so far.

Gerenuk, the statement of "first tell me why you think you can do it your way", really gets at why I raised this question in the first place. The problem I've always had with maths is that when asked a question like the one being analysed, I can only come up with answers which rely on performing the memorised arithmetic processes/substitution etc., rather than really providing the fundamental explanation.

Does anyone know if there is a field dedicated to studying the very fundamental nature of maths and how the axioms were established?

 

[chambershex]

And also, just one more question:

If the first question I posed can be shown to be true by reducing the binomial square down to an application of the distributive law, what can the following be reduced down to:

√(a + b) =/= √a + √b

So its pretty much the same question but instead of being squared the root is taken...

Really appreciate the help

 

[Mentallic]

And also, just one more question:

If the first question I posed can be shown to be true by reducing the binomial square down to an application of the distributive law, what can the following be reduced down to:

√(a + b) =/= √a + √b

So its pretty much the same question but instead of being squared the root is taken...

Really appreciate the help

It's really just an application of the same idea here in this thread. If x,y>0 (which is a given since we're dealing with the square roots of these numbers and only want to work in the domain with which this problem exists), then if x=y, then x²=y² since, by one of those laws (whcih I don't know the name of), if x=y then ax=ay, and let a=x=y to get x²=y².

Now let x=(a+b)^1/2 and y=a^1/2+b^1/2

 

[Dickfore]

If $x$ and $y$ are positive real numbers and you accept the axioms of Euclidean geometry, then this arguemt might appeal to you. See this figure:

[PLAIN]http://img413.imageshack.us/img413/1774/binomial.png
​

$(x + y)^{2}$ is the area of the large square. $x^{2}$ is the area of the square in the lower left corner and $y^{2}$ is the area of the square in the upper right corner. You see that their sum is smaller than the area of the big square. You need to include the areas of the two rectangles (which are identical) $2 x y$.

 

[alice22]

OK I know the answer has been given but it took me ages to draw this!

Anyone know of a better drawing package than MSpaint?

Edit looks like dickfore does!

 

[Dickfore]

Anyone know of a better drawing package than MSpaint?

Edit looks like dickfore does!

Actually, I used MSPaint. But, there are further options than the free-form line or brush line. Also, there is an option for typing characters.

 

[alice22]

Actually, I used MSPaint. But, there are further options than the free-form line or brush line. Also, there is an option for typing characters.

Thanks! Yours does look slightly better than mine!

I will try to do better in future!

It will lose the 'back of a fag packet' look though!

 

[alice22]

Actually, I have just been thinking, you could do a similar thing for (x+y) cubed.
I can't remember the the expansion for that but it would be a good exercise to do it
From a similar 3D diagram!

 

[Mentallic]

you could do a similar thing for (x+y) cubed.
I can't remember the the expansion for that

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

 

[chambershex]

It's really just an application of the same idea here in this thread. If x,y>0 (which is a given since we're dealing with the square roots of these numbers and only want to work in the domain with which this problem exists), then if x=y, then x²=y² since, by one of those laws (whcih I don't know the name of), if x=y then ax=ay, and let a=x=y to get x²=y².

Now let x=(a+b)^1/2 and y=a^1/2+b^1/2

Sorry I'm not really following your reasoning here. With the first problem I posed, you showed me how the left hand side (x+y)2 can be shown to be an application of the distributive law. However, even if the left hand side of this problem √(a + b) can be framed as (a+b)1/2, it isn't possible to make this of a form described in the distributive law is it? (I.e. of the form,
a x (b + c)).

Therefore, if it isn't possible to represent it as the distributive law axiom, which axiom can explain this problem? (I was under the impression, as Gerenuk alluded to before, that all mathematical expression can ultimately be explained by underlying axioms, which are accepted because they are self-evident).

 

[Gerenuk]

Gerenuk, the statement of "first tell me why you think you can do it your way", really gets at why I raised this question in the first place. The problem I've always had with maths is that when asked a question like the one being analysed, I can only come up with answers which rely on performing the memorised arithmetic processes/substitution etc., rather than really providing the fundamental explanation.

Then you are doing it correctly - it's not a problem. Never use any rule apart from the ones you have learned. Even the most brilliant mathematicians do that (well, maybe apart from Euler :) ) and only these kind of proofs are accepted in journals!
And if you see a "new" rule, then surely this rule was at some point derived from exclusively the more fundamental rules. If you have some intuition, then you still need get back to a well defined derivation from the basics.

 

[Gerenuk]

However, even if the left hand side of this problem √(a + b) can be framed as (a+b)1/2, it isn't possible to make this of a form described in the distributive law is it? (I.e. of the form,
a x (b + c)).

No. It can be framed as $(a+b)^{1/2}$ which is not the same as $(a+b)\cdot \frac12$. So the distributive law really applies only to multiplication (unless proved otherwise).

Therefore, if it isn't possible to represent it as the distributive law axiom, which axiom can explain this problem? (I was under the impression, as Gerenuk alluded to before, that all mathematical expression can ultimately be explained by underlying axioms, which are accepted because they are self-evident).

I wouldn't call it self-evident. It's rather about self-consistency. You are not allowed to define any rules that might violate others you have defined before.
Let's try that: We start by defining something that we call 1.
Then we need names for repeated 1s: 1+1=2, 1+1+1=3, 1+1+1+1=4, 1+1+1+1+1=5
So writing "2" is just an abbrevation for "1+1" - no more than that.
Now we also define 1*1=1 and the distributive law (a+b)*c=a*c+b*c
Now what is 2*2 now? Well, let's use our rules defined so far. We don't need anything else to define or know: 2*2=(1+1)*(1+1)=1*(1+1)+1*(1+1)=1*1+1*1+1*1+1*1=1+1+1+1=4
So without defining what 2*2 actually is, we can derive what it should be.
So it would have been a mistake to define 2*2=5. This would not be self-consistent with the previous rules.

In fact you can define any rules you want for your algebra. It would probably be better to use other symbols than the normal numbers for your new objects.
The important point is that only the ring structure (see wikipedia link; it's basically the algebra you know) is applicable to common problems in the real world.

Here is an example of an algebra where 3+6=5
http://en.wikipedia.org/wiki/Nimber
This algebra is useful for impartial games (games where both players have the same options to make a move) rather than the physical world.

 

[chambershex]

No. It can be framed as $(a+b)^{1/2}$ which is not the same as $(a+b)\cdot \frac12$. So the distributive law really applies only to multiplication (unless proved otherwise).


I wouldn't call it self-evident. It's rather about self-consistency. You are not allowed to define any rules that might violate others you have defined before.
Let's try that: We start by defining something that we call 1.
Then we need names for repeated 1s: 1+1=2, 1+1+1=3, 1+1+1+1=4, 1+1+1+1+1=5
So writing "2" is just an abbrevation for "1+1" - no more than that.
Now we also define 1*1=1 and the distributive law (a+b)*c=a*c+b*c
Now what is 2*2 now? Well, let's use our rules defined so far. We don't need anything else to define or know: 2*2=(1+1)*(1+1)=1*(1+1)+1*(1+1)=1*1+1*1+1*1+1*1=1+1+1+1=4
So without defining what 2*2 actually is, we can derive what it should be.
So it would have been a mistake to define 2*2=5. This would not be self-consistent with the previous rules.

In fact you can define any rules you want for your algebra. It would probably be better to use other symbols than the normal numbers for your new objects.
The important point is that only the ring structure (see wikipedia link; it's basically the algebra you know) is applicable to common problems in the real world.

Here is an example of an algebra where 3+6=5
http://en.wikipedia.org/wiki/Nimber
This algebra is useful for impartial games (games where both players have the same options to make a move) rather than the physical world.

Thanks for the detailed response. So using the same kind of process how can one prove that √(a + b) =/= √a + √b?

For the first problem I posed, (x+y)^2 =/= x^2 + y^2, I now understand (thanks to others) why it is so. For the left hand side, the expression (x+y)^2 is equivalent to (x+y)(x+y), and is thus of the form used in the distributive law, a x (b + c). Therefore the expression (x+y)(x+y) can simply be expanded following the rules of that law, resulting in x^2 + 2xy + y^2. The right hand side can simply be represented as x X x + y X y (because that is what exponents indicate). So the total equation can essentially be reduced down its basic terms using axiomatic rules. Consequently, on the both sides of the equation I can see that the terms x^2 + y^2 are common, however on the left hand side there is the additional term 2xy, which thus accounts for the difference.

Now for this equation √(a + b) =/= √a + √b, even if it is framed as (a+b)^1/2 =/= a^1/2 + b^1/2, the left side is not reducible to anything else. Therefore, how can one show using a similar process to the above how the left and right side are different (without using substitution)?

 

[Mentallic]

It's really just an application of the same idea here in this thread. If x,y>0 (which is a given since we're dealing with the square roots of these numbers and only want to work in the domain with which this problem exists), then if x=y, then x²=y² since, by one of those laws (whcih I don't know the name of), if x=y then ax=ay, and let a=x=y to get x²=y².

Now let x=(a+b)^1/2 and y=a^1/2+b^1/2

Sorry I'm not really following your reasoning here. With the first problem I posed, you showed me how the left hand side (x+y)2 can be shown to be an application of the distributive law. However, even if the left hand side of this problem √(a + b) can be framed as (a+b)1/2, it isn't possible to make this of a form described in the distributive law is it? (I.e. of the form,
a x (b + c)).

Just try what I asked you to do and you'll see where it leads you. Use the result proven here that (a+b)²=a²+2ab+b²

And since we're trying to show whether (a+b)^1/2=a^1/2+b^1/2

assume it's equal first. Let x=(a+b)^1/2, y=a^1/2+b^1/2

I've already shown you that if x=y, then x²=y².

Now substitute in and expand to see whether they're the same or not.

 

[chambershex]

Just try what I asked you to do and you'll see where it leads you. Use the result proven here that (a+b)²=a²+2ab+b²

And since we're trying to show whether (a+b)^1/2=a^1/2+b^1/2

assume it's equal first. Let x=(a+b)^1/2, y=a^1/2+b^1/2

I've already shown you that if x=y, then x²=y².

Now substitute in and expand to see whether they're the same or not.

Oh right, I get where you coming from now...

So yeh substitution results in:

a + b =/= a + 2√(ab) + b

Therefore the difference being 2√(ab).

 

[Gerenuk]

Thanks for the detailed response. So using the same kind of process how can one prove that √(a + b) =/= √a + √b?

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.

Now for this equation √(a + b) =/= √a + √b, even if it is framed as (a+b)^1/2 =/= a^1/2 + b^1/2, the left side is not reducible to anything else. Therefore, how can one show using a similar process to the above how the left and right side are different (without using substitution)?

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.

 

[Mentallic]

I don't think there is a useful way to write
$$\sqrt{a+b}=\sqrt{a}+\sqrt{b}+x$$

Sure there is, $$x=\sqrt{a+b}-\sqrt{a}-\sqrt{b}$$ :tongue2:

 

[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, $$x=\sqrt{a+b}-\sqrt{a}-\sqrt{b}$$ :tongue2:

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.