Question 485329


The proof that {{{sqrt(2)}}} is {{{irrational}}}:


 Let's {{{suppose}}} {{{sqrt(2)}}} were a {{{rational}}} number.  Then we can write it {{{sqrt(2)=a/b }}} where {{{a}}}, {{{b}}} are {{{whole}}} numbers, {{{b}}}{{{ not}}}{{{ zero}}}.  

We additionally make it so that this {{{a/b}}} is simplified to the lowest terms, since that can obviously be done with any fraction.

It follows that {{{2 = a^2/b^2}}},  or {{{ a^2 = 2 * b^2}}}.  So the square of {{{a}}} is an {{{even}}} {{{number}}} since it is {{{two}}} {{{times}}}{{{ something}}}.  

From this we can know that {{{a}}} itself is also an {{{even}}} number.  Why?  Because it can't be {{{odd}}}; if {{{a}}} itself was {{{odd}}}, then{{{ a * a}}} would be {{{odd}}}{{{ too}}}.  Odd number times odd number is always odd.  

-if {{{a}}} itself is an {{{even}}} number, then {{{a}}} is {{{2}}}{{{ times}}} some other whole number, or {{{a = 2k}}} where {{{k}}} is this other number.  We don't need to know exactly what {{{k}}} is; it won't matter.  Soon is coming the {{{contradiction}}}:

If we substitute {{{a = 2k}}} into the original equation {{{2 = a^2/b^2}}}, this is what we get:

{{{2= (2k)^2/b^2}}}

{{{2= 	4k^2/b^2}}}

{{{2*b^2= 4k^2}}}

{{{b^2	= 2k^2}}}

This means {{{b^2}}} is {{{even}}}, from which follows again that {{{b}}} itself is an {{{even}}}{{{ number}}}!

WHY is that a {{{contradiction}}}? Because we started the whole process saying that {{{a/b}}} is simplified to the lowest terms, and now it turns out that {{{a}}} and {{{b}}} would {{{both}}} be {{{even}}}. So {{{sqrt(2)}}} {{{CANNOT}}} be {{{rational}}}.

conclusion: {{{sqrt(2)}}} {{{HAS}}} to be {{{irrational}}}.