just how to prove: if \$a,b in chrischona2015.orgbb N\$, then \$a^1/b\$ is an creature or an irrational number? (13 answers)
I"m make the efforts to execute this evidence by contradiction. I understand I need to use a lemma to establish that if \$x\$ is divisible by \$3\$, then \$x^2\$ is divisible by \$3\$. The lemma is the straightforward part. Any type of thoughts? just how should I expand the proof because that this to the square root of \$6\$?

Say \$ sqrt3 \$ is rational. Then \$sqrt3\$ can be stood for as \$fracab\$, where a and also b have no common factors.

You are watching: Is the square root of 3 a rational number

So \$3 = fraca^2b^2\$ and \$3b^2 = a^2\$. Now \$a^2\$ have to be divisible by \$3\$, yet then so have to \$a \$ (fundamental organize of arithmetic). For this reason we have actually \$3b^2 = (3k)^2\$ and \$3b^2 = 9k^2\$ or even \$b^2 = 3k^2 \$ and also now we have a contradiction.

What is the contradiction?

The-Duderino through the way, the proof because that \$ sqrt6 \$ adheres to in the very same steps practically exactly. \$endgroup\$
suppose \$sqrt3\$ is rational, then \$sqrt3=fracab \$ for some \$(a,b)\$suppose we have actually \$a/b\$ in easiest form.eginalignsqrt3&=fracab\a^2&=3b^2endalignif \$b\$ is even, then a is also even in which instance \$a/b\$ is not in simplest form.if \$b\$ is odd then \$a\$ is additionally odd.Therefore:eginaligna&=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)+1endalignSince \$(n^2+n)\$ is an integer, the left hand side is even. Due to the fact that \$(3m^2+3m)\$ is one integer, the right hand side is odd and we have found a contradiction, because of this our hypothesis is false.

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A claimed equation \$m^2=3n^2\$ is a straight contradiction to the basic Theorem the Arithmetic, because when the left-hand next is expressed as the product of primes, there are evenly numerous \$3\$’s there, when there room oddly plenty of on the right.

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answered Sep 14 "14 at 5:05

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The number \$sqrt3\$ is irrational ,it can not be expressed as a ratio of integers a and also b. To prove that this statement is true, let united state Assume the it is rational and then prove it isn"t (Contradiction).

So the presumptions states that :

(1) \$sqrt3=fracab\$

Where a and also b room 2 integers

Now due to the fact that we desire to refuse our presumption in order to acquire our desired result, us must show that there space no such 2 integers.

Squaring both sides give :

\$3=fraca^2b^2\$

\$3b^2=a^2\$

(Note : If \$b\$ is odd then \$b^2\$ is Odd, climate \$a^2\$ is odd because \$a^2=3b^2\$ (3 time an weird number squared is odd) and Ofcourse a is odd too, because \$sqrtodd number\$ is also odd.

With a and also b odd, we deserve to say that :

\$a=2x+1\$

\$b=2y+1\$

Where x and also y need to be essence values, otherwise clearly a and b wont be integer.

Substituting these equations come \$3b^2=a^2\$ offers :

\$3(2y+1)^2=(2x+1)^2\$

\$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1\$

Then simplying and also using algebra us get:

\$6y^2 + 6y + 1 = 2x^2 + 2x\$

You should know 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 next is an even number. Why? (Similar Reason)

\$2x^2+2x\$ is even Always, and there is NO +1 favor there was 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 also b which meet the partnership = \$fracab\$ cannot it is in found.