I"m make the efforts to perform this proof by contradiction. I understand I need to use a lemma to create that if $x$ is divisible by $3$, climate $x^2$ is divisible by $3$. The lemma is the simple part. Any kind of thoughts? exactly how should I prolong the proof because that this come the square source of $6$?
Say $ \sqrt3 $ is rational. Then $\sqrt3$ deserve to be stood for as $\fracab$, whereby a and b have actually no typical factors.
You are watching: Is the square root of 3 rational or irrational
So $3 = \fraca^2b^2$ and $3b^2 = a^2$. Now $a^2$ should be divisible through $3$, yet then so should $a $ (fundamental organize of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and also now we have actually a contradiction.
What is the contradiction?
The-Duderino by the way, the proof for $ \sqrt6 $ follows in the exact same steps nearly exactly. $\endgroup$
suppose $\sqrt3$ is rational, then $\sqrt3=\fracab $ for part $(a,b)$suppose we have $a/b$ in simplest form.\beginalign\sqrt3&=\fracab\\a^2&=3b^2\endalignif $b$ is even, then a is likewise even in which case $a/b$ is not in simplest form.if $b$ is odd then $a$ is likewise odd.Therefore:\beginaligna&=2n+1\\b&=2m+1\\(2n+1)^2&=3(2m+1)^2\\4n^2+4n+1&=12m^2+12m+3\\4n^2+4n&=12m^2+12m+2\\2n^2+2n&=6m^2+6m+1\\2(n^2+n)&=2(3m^2+3m)+1\endalignSince $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the ideal hand side is odd and we have found a contradiction, thus our theory is false.
edited Mar 2 at 10:38
4555 bronze title
answer Sep 14 "14 at 4:24
71844 silver badges88 bronze badges
add a comment |
A claimed equation $m^2=3n^2$ is a direct contradiction come the fundamental Theorem the Arithmetic, since when the left-hand side is expressed as the product that primes, there room evenly many $3$’s there, while there are oddly numerous on the right.
answered Sep 14 "14 in ~ 5:05
57.2k44 gold badges5959 silver- badges119119 bronze title
add a comment |
The number $\sqrt3$ is irrational ,it can not be expressed as a ratio of integers a and also b. Come prove the this declare is true, let united state Assume that it is rational and then prove the isn"t (Contradiction).
So the presumptions states the :
Where a and also b space 2 integers
Now since we want to refuse our assumption in bespeak to get our wanted result, we must show that there space no such two integers.
Squaring both sides give :
(Note : If $b$ is odd then $b^2$ is Odd, then $a^2$ is odd due to the fact that $a^2=3b^2$ (3 time an strange number squared is odd) and also Ofcourse a is strange too, since $\sqrtodd number$ is also odd.
With a and b odd, we can say that :
Where x and y must be creature values, otherwise obviously a and also b wont be integer.
Substituting these equations to $3b^2=a^2$ provides :
$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$
Then simplying and also using algebra we get:
$6y^2 + 6y + 1 = 2x^2 + 2x$
You should know that the LHS is one odd number. Why?
$6y^2+6y$ is even Always, for this reason +1 to an even number provides an weird number.
The RHS side is an also number. Why? (Similar Reason)
$2x^2+2x$ is even Always, and there is NO +1 like there remained in the LHS to do it ODD.
See more: Is It Bad To Smoke Weed While On Antibiotics, Doctors Answer
There space no options to the equation because of this.
Therefore, integer values of a and also b which satisfy the partnership = $\fracab$ cannot it is in found.