I"m make the efforts to perform this evidence by contradiction. I understand I need to use a lemma to develop that if $x$ is divisible through $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any kind of thoughts? how should I extend the proof for this to the square root of $6$?
Say $ \sqrt3 $ is rational. Then $\sqrt3$ deserve to be stood for as $\fracab$, where a and also b have no common factors.
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So $3 = \fraca^2b^2$ and also $3b^2 = a^2$. Currently $a^2$ should be divisible by $3$, however then so need to $a $ (fundamental to organize of arithmetic). Therefore we have actually $3b^2 = (3k)^2$ and also $3b^2 = 9k^2$ or also $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, climate $\sqrt3=\fracab $ for some $(a,b)$suppose we have actually $a/b$ in most basic form.\beginalign\sqrt3&=\fracab\\a^2&=3b^2\endalignif $b$ is even, climate a is likewise even in which situation $a/b$ is not in simplest form.if $b$ is odd then $a$ is also 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 one integer, the left hand side is even. Due to the fact that $(3m^2+3m)$ is an integer, the best hand side is odd and we have discovered a contradiction, as such our hypothesis is false.
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answer Sep 14 "14 at 4:24
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A an alleged equation $m^2=3n^2$ is a straight contradiction to the fundamental Theorem that Arithmetic, due to the fact that when the left-hand next is expressed as the product the primes, there are evenly plenty of $3$’s there, when there are oddly plenty of on the right.
answered Sep 14 "14 in ~ 5:05
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The number $\sqrt3$ is irrational ,it cannot be expressed together a proportion of integers a and also b. To prove that this declare is true, let us Assume the it is rational and then prove that isn"t (Contradiction).
So the presumptions states that :
Where a and b room 2 integers
Now because we desire to refuse our assumption in stimulate to acquire our preferred result, us must show that there room no such 2 integers.
Squaring both sides give :
(Note : If $b$ is odd then $b^2$ is Odd, climate $a^2$ is odd since $a^2=3b^2$ (3 times an weird number squared is odd) and also Ofcourse a is weird too, due to the fact that $\sqrtodd number$ is also odd.
With a and b odd, we deserve to say the :
Where x and also y should be integer values, otherwise clearly a and b no be integer.
Substituting this equations to $3b^2=a^2$ provides :
$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$
Then simplying and using algebra us get:
$6y^2 + 6y + 1 = 2x^2 + 2x$
You should understand that the LHS is an odd number. Why?
$6y^2+6y$ is even Always, therefore +1 come an even number gives an strange number.
The RHS side is an even 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.
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There room no options to the equation due to the fact that of this.
Therefore, integer worths of a and b which satisfy the connection = $\fracab$ cannot it is in found.