Question 91198
<pre>
Let us suppose that {{{root(3,7)}}} is a rational number. Then we can write {{{root(3,7)}}} = {{{m/n}}}, where {{{m}}} and {{{n}}} are whole numbers. 

We can assume that the numbers {{{m}}} and {{{n}}} have no common divisor. Otherwise, we can cancel this common divisor 

in the numerator and denominator. 

Raise both sides of the equality {{{root(3,7)}}} = {{{m/n}}} in the degree 3. You will get {{{m^3/n^3}}} = {{{7}}}. Hence, {{{m^3}}} = {{{7*n^3}}}.


The right side of the last equality is divisible by {{{7}}}. Hence, its left side {{{m^3}}} is divisible by {{{7}}} also. 
Since {{{m^3}}} is divisible by {{{7}}}, the whole number {{{m}}} itself is divisible by {{{7}}}. 

So, we can write {{{m}}} = {{{7*m[1]}}}, where {{{m[1]}}} is another whole number. 

Now, substitute {{{m}}} = {{{7*m[1]}}} into {{{m^3}}} = {{{7*n^3}}}. It gives {{{7^3*m[1]^3}}} = {{{7*n^3}}}. Cancel both sides by {{{7}}}. You will get {{{7^2*m[1]^3}}} = {{{n^3}}}. 

The left side of the last equality is divisible by {{{7}}}. Hence, its right side {{{n^3}}} is divisible by {{{7}}} also. 
Since {{{n^3}}} is divisible by {{{7}}}, the whole number {{{n}}} itself is divisible by {{{7}}}.

We just got a contradiction. We assumed that {{{root(3,7)}}} = {{{m/n}}}, where {{{m}}} and {{{n}}} are whole numbers with no common divisor, and the chain 
of arguments led us to the conclusion that both the integers {{{m}}} and {{{n}}} have the common divisor {{{7}}}. 

This contradiction proves that the original assumption was wrong, regarding {{{root(3,7)}}} as a rational number. Hence, {{{root(3,7)}}} is irrational. 
The proof is completed. 


In the proof we used many times this property of the number 7: if 7 divides the product of two integers m and n, then 7 divides 
at least one of the integers. It is the common property of any prime number in the ring of integer numbers.
</pre>