What is the reason for commutative property during multiplication of real numbers This may seem like a really stupid question, but I am unable to rationalize with myself as to why the commutative property exists when multiplying 2 real number.
Take for example:
$2 * 5 = 10$
This actually means that if we add $2$ $5$ times we will get $10$.
But its kind of amazing when we can say for sure that if we add $5$ $2$ times we will also get $10$.
What is the reason for this property, I know I may be over-thinking this, but I can't understand intuitively why this happens.
I know that $5$ actually "contains" $2$ but how does that guarantee commutative property ? 
For example matrix multiplication is not commutative, yet real number multiplication is.
Am I just overthinking something simple? 
Need some clarity.
 A: Is $17 \times 19$ equal to $19 \times 17$? This is not obvious, until we draw a certain picture, and then it does become quite obvious. 
Draw a rectangular array of dots with 17 rows and 19 columns. If we group the dots row by row, then we have 17 groups of 19. On the other hand, if we group the dots column by column, then we have 19 groups of 17. Thus, 17 of 19 is the same thing as 19 of 17.
A:  * * * * *
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In this picture there are five columns of two stars each, so in total there are $2+2+2+2+2$ stars.
On the other hand, there are also two rows of five stars each, so in total there are $5+5$ stars.
Since the number of stars has to be the same no matter how we count them -- they're the same stars! -- it must necessarily be that $2+2+2+2+2=5+5$.
The fact that if you count the same things in different orders you will get the same number is arguably the most fundamental property of "counting". It is a practical experience that ought to be supported by plenty of actual examples in elementary school. I doubt it can be reduced to something that feels more fundamental than counting itself does.
A: For thousands of years before if could be formally proven, the commutativity of addition and multiplication was held to be "common sense" based on experience with countless examples.  Today, it could even be used as a kind of test of any formal definition of numbers (natural or real), addition and multiplication. If you could not prove these and other "common sense" properties using these definitions, they would  clearly be inadequate. 
A: You can think of it like this maybe: 2 times 5 is the total size of 5 sets with 2 elements each, whereas 5 times 2 is the total size of 2 sets with 5 elements each. To see that these represent the total same number of elements, we see that we can match up the first item of the third set in the first product with the third item of the first set in the second product, and the second item of the fourth set in the first product with the fourth item of the second set in the second product, etc. If we can pair up the items in two groups one by one, they must have the same size. 
A: Note:- I am providing this answer using the example. If any could generalise it, i shall be grateful to him. :)
Prove that $14×11=11×14$. (I suppose that you know mxn is n added m times)
Proof-: $11×14=(11×11)+(11×3)$.....(1)
Let $3×11=11×3$.
So, the eq.(1) becomes =$(11×11)+(3×11)=14×11$. Proved.
But, here we supposed that $3×11=11×3$. We need to prove the same.
So, consider $11×3=(3×3)+(8×3)$....(2)
Let $8×3=3×8$. 
So, (2) becomes $11×3=(3×3)+(3×8)=3×11$. Proved. 
But in the latter we considered that $8×3=3×8$. So, we need to prove the same. In the similar manner, we prove that $8×3=3×8$, but supposing that $3×5=5×3$. So, we need to prove this now. We shall prove this in the similar manner supposing that $3×2=2×3$. Now, we need to prove this. We prove this in the similar manner and at an intermediate step, we suppose that $2×1=1×2$. But look at this. Do you believe in $n×1$ i.e. n  equals $1×n$, that is $1$ added n times. So, you would always end up supposing $c×1=1×c$, which is true by the definition of natural numbers. Therefore, we proved that $14×11=11×14$. Now, its your turn to generalise this for the following statement.
For some +ve integers $a,b$,$a×b=b×a$.
