How to prove: if $a,b in

adechworld.combb N$, then $a^{1/b}$ is an integer or an irrational number? (13 answers)

I”m trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of $6$?

Say $ sqrt{3} $ is rational. Then $sqrt{3}$ can be represented as $frac{a}{b}$, where a and b have no common factors.

You are watching: Square root of 3 is irrational

So $3 = frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

What is the contradiction?

The-Duderino By the way, the proof for $ sqrt{6} $ follows in the same steps almost exactly. $endgroup$

suppose $sqrt{3}$ is rational, then $sqrt{3}=frac{a}{b} $ for some $(a,b)$suppose we have $a/b$ in simplest form.egin{align}sqrt{3}&=frac{a}{b}\a^2&=3b^2end{align}if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.if $b$ is odd then $a$ is also odd.Therefore:egin{align}a&=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)+1end{align}Since $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

Share

Cite

Follow

edited Mar 2 at 10:38

bilibraker

4755 bronze badges

answered Sep 14 “14 at 4:24

qwerty314qwerty314

71844 silver badges88 bronze badges

$endgroup$

5

Add a comment |

10

$egingroup$

A supposed equation $m^2=3n^2$ is a direct contradiction to the Fundamental Theorem of Arithmetic, because when the left-hand side is expressed as the product of primes, there are evenly many $3$’s there, while there are oddly many on the right.

Share

Cite

Follow

answered Sep 14 “14 at 5:05

LubinLubin

57.4k44 gold badges6060 silver badges119119 bronze badges

$endgroup$

Add a comment |

4

$egingroup$

The number $sqrt{3}$ is **irrational** ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is **rational** and then prove it isn”t (Contradiction).

So the Assumptions states that :

**(1)** $sqrt{3}=frac{a}{b}$

Where a and b are 2 integers

**Now since we want to disapprove our assumption in order to get our desired result, we must show that there are no such two integers.**

See more: Net Ionic Equation Of Acetic Acid And Sodium Hydroxide, Sodium Hydroxide (Aq) + Acetic Acid (Aq) →

Squaring both sides give :

$3=frac{a^2}{b^2}$

$3b^2=a^2$

(**Note** : If $b$ is **odd** then $b^2$ is Odd, then $a^2$ is odd because $a^2=3b^2$ (3 times an odd number squared is odd) and Ofcourse a is odd too, because $sqrt{odd number}$ is also odd.

With **a** and **b** odd, we can say that :

$a=2x+1$

$b=2y+1$

Where x and y must be integer values, otherwise obviously a and b wont be integer.

Substituting these equations to $3b^2=a^2$ gives :

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

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

Then simplying and using algebra we 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, so +1 to an even number gives an ODD number.

The RHS side is an even number. Why? (Similar Reason)

$2x^2+2x$ is **even** Always, and there is NO +1 like there was in the LHS to make it ODD.

See more: What Is 15 Is What Percent Of 35 ? = 0 15 Is What Percent Of 35

There are no solutions to the equation because of this.

Therefore, integer values of a and b which satisfy the relationship = $frac{a}{b}$ **cannot** be found.