Proof of Distributive Rule: Why a(b+c)=ab+ac

[Mentallic]

I am curious to see a proof as to why a(b+c)=ab+ac.
Also is this a relatively new rule in the world of mathematics because I remember hearing somewhere that expressing quadratics and a product of two factors wasn't done till quite recently (how recent this was I am not sure of).

Thanks.

 

[csopi]

I suppose you mean that a,b,c are elements of a ring. If so, then a(b+c)=ab+ac by the definition of a ring.

 

[Mentallic]

Is there no other reason? Because I don't know what rings are and the fact that they work for all real numbers seems like it should have a decent proof behind it - not a definition in some field of maths.

 

[csopi]

Real numbers are not present in nature, they are abstract creations of mankind. In fact, only positve integers (and maybe the zero) are there "as is". If you want to create negative numbers, not to mention rational numbers and the real numbers in a rigorous mathematic way, you have to make a lot of axioms and algebraic definitions (such as additive and multiplicative inverse, equivalence class, quotient space, limit point ...). Of course, they are choosed "properly", so that they satisfy our intuitions we acquired by multiplying and adding pos. integers, but you have to see, that real numbers are our creations. The only reason, they seem so natural is that you've got used to them (since you've been calculating with them since your early childhood).

The equation is trivial for positive integers, because for them the multiplication is just a short notation for multiple additions. But when you go further, the only reason it remains true is that you choosed the definitions properly...

 

[Gerenuk]

Is there no other reason? Because I don't know what rings are and the fact that they work for all real numbers seems like it should have a decent proof behind it - not a definition in some field of maths.

It's the other way round. Number products are defined by the distributive rule!
Here is a simple reasoning:
We start with 1; We define that 1*1=1
Now we need a new symbol for 1+1; so we invent 2=1+1
Next we need a symbol for 1+1+1 and we invent 3=1+1+1
4=1+1+1+1
5=1+1+1+1+1

OK, but what's 2*2 now?
We can *proof* that it has to be 4, so we *dont have to invent* a new symbol for 2*2:
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
You see that basically the distributive law enables us to find what products should be equal to.

 

[zgozvrm]

Try this:

We know that ab = b + b + b + ... + b ("b" added to itself "a" times) by definition of multiplication

To make things a bit easier to follow, let's enumerate each instance of "b" so we can keep track. We'll say that ab = b_1 + b_2 + ... + b_{a-1} + b_a[/tex]<br /> <br /> Therefore, it is also true that a(b + c) = (b + c)_1 + (b + c)_2 + ... + (b + c)_{a-1} + (b + c)_a[/tex]&lt;br /&gt; &lt;br /&gt; This is really (b_1 + c_1) + (b_2 + c_2) + ... + (b_{a-1} + c_{a-1}) + (b_a + c_a)[/tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Which can be rearranged with the associative rule into (b_1 + b_2 + ... + b_{a-1} + b_a) + (c_1 + c_2 + ... + c_{a-1} + c_a)[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; By definition of multiplication, this is ab + ac

 

[Mentallic]

While some have said it is a definition, what zgozvrm has posted seems like a valid proof of the distributive law by only using the definitions of what multiplication really is. A definition can't be proven by other definitions can it? I mean, isn't that what the idea of an axiom is?

 

[zgozvrm]

A definition can't be proven by other definitions can it? I mean, isn't that what the idea of an axiom is?

Yes, you have to start somewhere...
At some point multiplication was defined as repeated addition.
Addition was defined prior to that. (We had to have some way of describing what it means to have "a" items and increasing that amount by "b" items).

Multiplication made it easier to deal with larger numbers, arrays, etc.

If everyone agrees on the rules, we can use those rule to create/prove new ones, as long as the original rules aren't broken.

In any formal proof, each step is justified by a definition, theorem, postulate, etc. that has been previously shown to be true. So each proof becomes a building block for other proofs.

There is no need to waste time "reinventing the wheel."


What you can't do, is use a definition to define itself like:
"Addition is the process of adding 2 or more numbers together, resulting in a single sum."

 

[Mentallic]

Right, so isn't the distributive rule a consequence of the multiplication and addition definitions as you've shown, rather than a definition in itself?

I mean when they were creating the foundations of mathematics, they would start with defining addition and such, and then defining multiplication from addition, but rather than defining the distributive rule, they proved the distributive rule using the already known definitions.

 

[zgozvrm]

By the way, axioms are the building blocks of mathematical theorems.
They can't be proven, they are just accepted to be true; they are self-evident.

The fact x = x is an axiom. There is no mathematical proof of this, yet we all know it to be true. (Given that we know and accept what the concept of equality is).

 

[zgozvrm]

Right, so isn't the distributive rule a consequence of the multiplication and addition definitions as you've shown, rather than a definition in itself?

I mean when they were creating the foundations of mathematics, they would start with defining addition and such, and then defining multiplication from addition, but rather than defining the distributive rule, they proved the distributive rule using the already known definitions.

Absolutely!

 

[Mentallic]

Yes, thank you And that proof was very elegant, I appreciate it.

 

[jgens]

Try this:

We know that ab = b + b + b + ... + b ("b" added to itself "a" times) by definition of multiplication

To make things a bit easier to follow, let's enumerate each instance of "b" so we can keep track. We'll say that ab = b_1 + b_2 + ... + b_{a-1} + b_a[/tex]<br /> <br /> Therefore, it is also true that a(b + c) = (b + c)_1 + (b + c)_2 + ... + (b + c)_{a-1} + (b + c)_a[/tex]&lt;br /&gt; &lt;br /&gt; This is really (b_1 + c_1) + (b_2 + c_2) + ... + (b_{a-1} + c_{a-1}) + (b_a + c_a)[/tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Which can be rearranged with the associative rule into (b_1 + b_2 + ... + b_{a-1} + b_a) + (c_1 + c_2 + ... + c_{a-1} + c_a)[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; By definition of multiplication, this is ab + ac

&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; What would it mean to add a number to itself, say, 2&amp;amp;lt;sup&amp;amp;gt;1/2&amp;amp;lt;/sup&amp;amp;gt; times?

 

[zgozvrm]

I assume you meant 2.5 (rather than 2^{1/2} = \sqrt{2}[/tex]) times?<br /> <br /> You would first have to introduce/define fractions and division.

 

[zgozvrm]

I assume you meant 2.5 (rather than 2^{1/2} = \sqrt{2}[/tex]) times?<br /> <br /> You would first have to introduce/define fractions and division.

<br /> <br /> Then you could show that 2 \frac{1}{2} = \frac{5}{2}[/tex]&lt;br /&gt; &lt;br /&gt; Then you would multiply the numerators (using the definition of multiplication)&lt;br /&gt; and multiply the denominators (again, using the definition of multiplication)&lt;br /&gt; &lt;br /&gt; The result would be the product of the numerators over the product of the denominators.&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; ... but I&amp;#039;m sure you already knew this (just trying to be &amp;quot;smart&amp;quot;)!

 

[slider142]

This is a good motivation for defining the distributive property for real numbers, but it does not work as a theorem in general. For example, how would your example work for Euler's number multiplied by the constant pi or irrationals in general?

 

[zgozvrm]

Multiplication is still repeated addition.

e.g. 4 * \pi[/tex] can never be exactly determined. But, we can deternime that value to a certain number of places. <br /> <br /> Let&#039;s use \pi[/tex] to 2 places: 3.14&lt;br /&gt; We can show this as 3 + 0.1 + 0.04&lt;br /&gt; If we were to multiply this by 4, we can show that 4 * 3.14 = 4 * (3 + 0.1 + 0.04) = 4*3 + 4*0.1 + 4*0.04 &lt;br /&gt; &lt;br /&gt; You can continue \pi[/tex] out as far as you like and the distributive property still holds true. The only problem is that we have to stop at some point because we have yet to find an exact value of \pi[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; This can be shown for all types of numbers, so, yes it does work as a theorem in general.

 

[zgozvrm]

it does not work as a theorem in general

In fact, it is a theorem, in general!

I may not have shown it to work "in general" but I also didn't define fractional, irrational or complex numbers and how to multiply them.

I believe the distributive property is generally introduced to math students before they learn about numbers other than integers. When they learn about other types of numbers, that is when the elementary laws of mathematics (associative, distributive, commutative) are shown to still work, not the other way around.

 

[Mentallic]

Multiplication is still repeated addition.

e.g. 4 * \pi[/tex] can never be exactly determined. But, we can deternime that value to a certain number of places. <br /> <br /> Let&#039;s use \pi[/tex] to 2 places: 3.14&lt;br /&gt; We can show this as 3 + 0.1 + 0.04&lt;br /&gt; If we were to multiply this by 4, we can show that 4 * 3.14 = 4 * (3 + 0.1 + 0.04) = 4*3 + 4*0.1 + 4*0.04 &lt;br /&gt; &lt;br /&gt; You can continue \pi[/tex] out as far as you like and the distributive property still holds true. The only problem is that we have to stop at some point because we have yet to find an exact value of \pi[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; This can be shown for all types of numbers, so, yes it does work as a theorem in general.

&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; This method wouldn&amp;amp;amp;#039;t be able to show that \sqrt{2} multiplied by itself is in fact 2.

 

[zgozvrm]

This method wouldn't be able to show that \sqrt{2} multiplied by itself is in fact 2.

It's not supposed to!

The fact that \sqrt{2} \times \sqrt{2} = 2[/tex] has NOTHING to do with the distributive property!<br /> <br /> <br /> It also doesn&#039;t show that 4/2 = 2, or that a rectangle has 4 sides, or that the volume of a sphere is equal to \frac{4}{3}\pi r^3[/tex]

 

[zgozvrm]

This method wouldn't be able to show that \sqrt{2} multiplied by itself is in fact 2.

Although \sqrt{2} \times \sqrt{2}[/tex] is a multiplication, you are actually dealing with exponents and square roots here:<br /> <br /> \sqrt{2} \times \sqrt{2} = (\sqrt{2})^2<br /> <br /> Now, you <i>can</i> multiply them using traditional methods (i.e. repeated addition) but then you&#039;d be dealing with approximations because \sqrt{2}[/tex] is just another irrational number (approximately equal to 1.414213). So my explanation of how to multiply a number by \pi[/tex] holds for \sqrt{2}[/tex], as well!

 

[zgozvrm]

While the proof I gave uses multiplication and division, I don't have to show that multiplication and division work for all types of numbers. Rather, it's the other way around!

No matter WHAT kind of number you're talking about, multiplication comes down to repeated addition (look it up).

I merely offered a way to show that the distributive property is valid.

 

[Tac-Tics]

The distributive property of natural numbers is axiomatic.

All other distributive properties (of integers, rationals, reals) follow from this (and the other axioms of arithmetic).

 

[zgozvrm]

The distributive property of natural numbers is axiomatic.

All other distributive properties (of integers, rationals, reals) follow from this (and the other axioms of arithmetic).

No, the distributive property of natural numbers is not "axiomatic," as you say. It is not an axiom because it can be proven mathematically and it is not self-evident. (Look up the definition of a mathematical axiom.)

 

[Tac-Tics]

You can choose whatever axioms you want as long as they don't lead to contradiction.

And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic.

You can, of course, be clever and come up with an alternative set of axioms where distributivity is a theorem, but it's such a fundamental property, any proof would be uninteresting to the OP.

 

[zgozvrm]

You can choose whatever axioms you want as long as they don't lead to contradiction.

And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic.

You can, of course, be clever and come up with an alternative set of axioms where distributivity is a theorem, but it's such a fundamental property, any proof would be uninteresting to the OP.

Again, the distributive property is not an axiom since it is not self-evident. The fact that x = x is an axiom, because it is self-evident and can't be proven formally.
After all it is called the distributive property, not the distributive axiom!



And, I disagree ... the proof is, in fact interesting to the OP, since that is exactly what they were asking for!

I am curious to see a proof as to why a(b+c)=ab+ac.

 

[Tac-Tics]

After all it is called the distributive property, not the distributive axiom!

The property in question is included as one of the ring axioms. It's true by definition, and thus satisfies the criteria of axiom-hood. (Being "self-evident" has no bearing).

 

[Hurkyl]

Multiplication is still repeated addition.

e.g. 4 * \pi[/tex] can never be exactly determined.

<br /> 4\pi seems rather exact to me. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" /><br /> <br /> <br /> <br /> Anyways, this thread is getting too silly. You&#039;re viewing things far too narrowly, zgozvrm.

 

[zgozvrm]

The property in question is included as one of the ring axioms. It's true by definition, and thus satisfies the criteria of axiom-hood. (Being "self-evident" has no bearing).

The distributive property is hardly true "by definition." The definition of the distributive property is a(b + c) = ab + ac, but that's not immediately evident, and so must be shown to work (i.e. proven) before it can be used.

Axioms are used in proving that the distributive property works, but that doesn't make the distributive property, itself, an axiom. Otherwise every mathematical theorem, property, etc. would be considered an axiom.

An axiom is a building block upon which properties, theorems, etc. are proven.

 

[zgozvrm]

4\pi seems rather exact to me.



Anyways, this thread is getting too silly. You're viewing things far too narrowly, zgozvrm.

Really? So you can exactly determine the value of \pi[/tex]?<br /> You must publish your results. Mathematicians have been trying to find that value for centuries!

 

[zgozvrm]

4\pi seems rather exact to me.



Anyways, this thread is getting too silly. You're viewing things far too narrowly, zgozvrm.

Besides, \pi[/tex] is merely a symbol representing a value; a value which cannot be exactly determined.<br /> <br /> And, no, I&#039;m not viewing things too narrowly. Rather, others are trying to dispute a known fact, perhaps in an attempt to try and get me to print a &quot;perfect&quot; widely-acceptable definition.<br /> <br /> The OP was looking for a way to prove the distributive property, and I gave him one. I gave him an algebraic proof that works. Although I didn&#039;t prove that it works for irrational numbers, or complex numbers, or integers, or fractional numbers, etc. specifically, my proof can be used to show that it does work for all those types of numbers (using other known axioms, properties, etc.) But that wasn&#039;t the point of the original question, now was it?<br /> <br /> By the way the OP was happy with the solution I gave, so why everyone has to pick it apart afterward, is beyond me.

 

[Tac-Tics]

Really? So you can exactly determine the value of \pi[/tex]?<br /> You must publish your results. Mathematicians have been trying to find that value for centuries!

<br /> <br /> You&#039;re trolling the wrong forum if you think pi doesn&#039;t have a exact and well-understood value.

 

[Hurkyl]

Really? So you can exactly determine the value of \pi[/tex]?<br /> You must publish your results. Mathematicians have been trying to find that value for centuries!

<br /> What&#039;s to determine? Even the ancient Greeks knew the exact value of pi, and the entire decimal form has been known for centuries, at least.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tac-Tics" data-source="post: 2891273" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tac-Tics said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic. </div> </div> </blockquote>The Peano axioms only talk about the successor operation.

 

[Hurkyl]

And, no, I'm not viewing things too narrowly. Rather, others are trying to dispute a known fact, perhaps in an attempt to try and get me to print a "perfect" widely-acceptable definition.

Yes, you are. You have your own pet idea about words like "value" or "axiom" or "property", and you complain whenever anybody doesn't conform to your narrow world view.

 

[zgozvrm]

More often than not, this forum seems to be a place where people try and disprove those who attempt to help others rather than offering up additional information to help support the answer. I guess they were too small to stand up for themselves on the playground, and now get their jollies "bullying" people here.

We all know that the distributive property is indeed a property (not an axiom) and that it works. The OP was looking for a way to prove it (perhaps to help explain to someone else that it works in all situations) He wasn't asking me to prove that it is an axiom or that it is a property. I tried to help him out. No need to question that multiplication is repeated addition ("What would it mean to add a number to itself, say, 21/2 times?") or that my example doesn't work "for Euler's number multiplied by the constant pi or irrationals in general" (when, in fact, it does ... I just didn't show it).

 

[Tac-Tics]

The Peano axioms only talk about the successor operation.

Yeah, I was mistaken here.

I conflated peano's axioms with the axioms of TNT in Hofstadter's axiomatization of arithmetic (which is an example of an axiomatic system where distributivity is in fact an axiom).

 

[zgozvrm]

Yes, you are. You have your own pet idea about words like "value" or "axiom" or "property", and you complain whenever anybody doesn't conform to your narrow world view.

First of all, I'm not complaining.

Secondly, is anyone here really disputing that, for instance, multiplication is not repeated addition. Or, that that the distributive property is NOT an axiom? Or, that you cannot write the value of \pi[/tex] <i>exactly</i> (which is why we use a symbol to represent it)?

 

[zgozvrm]

What's to determine? Even the ancient Greeks knew the exact value of pi, and the entire decimal form has been known for centuries, at least.

By "the entire decimal form," you are referring to 3.141592654... , correct?

Can you show me where I can find reference to this fact? It may be an exact number, but as far as I know, the "decimal part" (0.141592654...) has never been shown to repeat at any point. And therefore, cannot even be shown as a fractional number (an exact ratio). So, until that happens, how can we say we know the exact decimal value?

 

[Hurkyl]

So, until that happens, how can we say we know the exact decimal value?

By proof. A decimal number is a function from N to the set {0,1,2,3,4,5,6,7,8,9,.} with exactly one occurrence of '.'; while somewhat cumbersome to write and to compute, it's quite straightforward from the Taylor series for arctan(1) centered at 0 together with the Taylor remainder theorem. Much more efficiently computable forms are known these days.

 

[zgozvrm]

By proof. A decimal number is a function from N to the set {0,1,2,3,4,5,6,7,8,9,.} with exactly one occurrence of '.'; while somewhat cumbersome to write and to compute, it's quite straightforward from the Taylor series for arctan(1) centered at 0 together with the Taylor remainder theorem. Much more efficiently computable forms are known these days.

Yes, that is what a decimal number is. But that doesn't mean that we know the exact value of \pi[/tex]. If we did, we&#039;d be able to express it in terms of the ratio of 2 whole numbers.<br /> <br /> You said it yourself... &quot;Much more efficiently computable forms are known these days.&quot;<br /> The value of \pi[/tex] has been computed to many more places (and more efficiently), but it has yet to be exactly determined.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;Mentallic&quot; data-source=&quot;post: 2887546&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; Mentallic said: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; Yes, thank you &lt;img src=&quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png&quot; class=&quot;smilie smilie--emoji&quot; loading=&quot;lazy&quot; width=&quot;64&quot; height=&quot;64&quot; alt=&quot;:smile:&quot; title=&quot;Smile :smile:&quot; data-smilie=&quot;1&quot;data-shortname=&quot;:smile:&quot; /&gt; And that proof was very elegant, I appreciate it. &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;Can we just accept that the OP has received the answer he was looking for, and stop nit-picking?

 

[Hurkyl]

Yes, that is what a decimal number is. But that doesn't mean that we know the exact value of \pi[/tex]. If we did, we&#039;d be able to express it in terms of the ratio of 2 whole numbers.

<br /> This is only true for your narrow view of the word &quot;value&quot;.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You said it yourself... &quot;Much more efficiently computable forms are known these days.&quot; </div> </div> </blockquote>Huh? What does the ability to compute with a function have to do with whether or not we know the function itself?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Can we just accept that the OP has received the answer he was looking for, and stop nit-picking? </div> </div> </blockquote>The &quot;nitpicking&quot; has continued because you have dug in your heels and defend an indefensible position. The original &quot;nitpick&quot; about your original proof (it was by no means clear that the OP was interested specifically in how to prove the distributive law for the natural numbers from the specific premises you opted to use) comes from the striking resemblance of your argument to one of the standard crackpot stances that only multiplication by an integer is meaningful.

 

[zgozvrm]

Can we just accept that the OP has received the answer he was looking for, and stop nit-picking?

Apparently not.

 

[zgozvrm]

Done.

 

[mathwonk]

one of the earliest treatments of this as a theorem is in euclid, where the quantities are undefined measures of size of rectangles, and he then proves this by decomposing the rectangle with height a, and length (b+c), into two rectangles, of lengths b and c.

 

[Mentallic]

This seems to have gotten a little out of hand. It's exactly what I was looking for, a proof to show that a(b+c)=ab+ac and the proof that zgozvrm gave was enough to satisfy this for me. I didn't really mind about what restrictions were placed on the constants but mathwonk's little geometric proof extends this to all real numbers. That was nice of you as well.

This is what I wanted to see, not pages of bickering. Thank you.

And by the way, the answer has been kind of lost in the mess, but if the distributive property can be proven by more simple definitions (or axioms) of multiplication and addition, is the distributive property itself an axiom?

 

[l'Hôpital]

I think it really depends on what you define axioms. I mean, do we take "+" for granted? What are the axioms?

If we work with, say, the Peano Axioms (and the elements are natural numbers), then it's a theorem. In particular, it can be proven rigorously using induction/the successor function.

 

[Mentallic]

Yes you're right. It all just depends on how you define axioms and since I have a primitive understanding of them - I assumed they mean "starting off points" in maths.

 

[Gerenuk]

To believe zgozvrm's "proof" is naive, because he showed it for integer numbers only.
There is a reason why millions of mathematicians in the world will tell you that the distributive law is an axiom. And the majority of people here agree.
zgozvrm should first read how to write down rigorous proofs, before claiming he can generalize his toy example to irrational numbers.

EDIT: but reading all the lasts posts now I see that there are enough clever people here who can explain it better than me.

 

[Mentallic]

There is a reason why millions of mathematicians in the world will tell you that the distributive law is an axiom. And the majority of people here agree.

Right, so what is it? I'm not disputing this but it just seems logical for the moment that since proofs were given for this rule, it isn't an axiom.

 

[Gerenuk]

Right, so what is it? I'm not disputing this but it just seems logical for the moment that since proofs were given for this rule, it isn't an axiom.

You have think in the "pedantic" way a mathematician does and there every little bit is important: Strictly speaking the proof that was given *presupposes* a definition of multiplication. Basically it _assumes_ a*b=(a-1)*b+b. That's the hidden assumption there. It's reasonable from the real world perspective.
But it cannot be generalized to irrational numbers, so you have not proven the distributive law for general numbers! Therefore mathematicians take another approach, which incidently includes the simple integer case.