#10**+8 **

Here's another proof:

Let's suppose that there are two positive, relatively-prime integers, call them a and b, such that a/b = √5 with a > b.

Multiply both sides by b, so we have

a = b√5

Now square both sides

a^{2} = 5b^{2}

And note that we can write 5b^{2} as 4b^{2} + b^{2 } Now, subtract b^{2} from both sides....so we have...

a^{2} - b^{2} = 4b^{2}

Now, note that the right side is even, and suppose that a and b are both even. But, this is impossible since they are relatively prime and thus, have no factors in common.

And neither can one be odd an the other even, because the left side would be odd, but the right side is even.

The only possibility left is that both are odd. But this can't be so for the following reason: Factor the left side as (a + b) (a - b). And if a and b are odd, then (a + b) is even and so is (a - b) - (remember that a > b, so a - b > 0).

So we have

(a + b) ( a - b) = 4b^{2} Now divide both sides by 4 ....... and we can write

[(a + b)/2] [(a - b)/2] = b^{2}

Now, if (a-b)/2 is odd, then (a+b)/2 is even ..... and vice-versa. To see this, let (a - b)/2 = 2k + 1. Then a - b = 2(2k + 1) .... Add b to both sides..... a = 2(2k + 1) + b....Now add b again to both sides and we have a + b = 2(2k +1) + 2b.......Now, divide by 2 on both sides, and we have (a+b)/2 = (2k + 1) + b, and both sides are even. Likewise, it can be proved that if (a +b)/ 2 is odd, then (a-b)/2 is even.

But, an even times an odd is always even...... but b^{2} is odd.

Thus......a and b aren't both even or both odd, and neither are they of different parity. and nothing else can be possible. Thus, our assumption that there exists two positive, relatively-prime integers such that a/b = √5 must be false. And thus, √5 must be irrational.

CPhill Aug 31, 2014

#1**0 **

Yep, if you can't write it as a fraction with only integers, it's irrational.

Will85237 Aug 31, 2014

#2**0 **

Maybe it is thousands of digits over thousands of digits, then terminates.

How do we know we cannot write it that way?

Guest Aug 31, 2014

#3**+5 **

Yes the square root of 5 is an irrational number.

Here is a proof (by contradiction)

$$I will assume that $\sqrt5$ is a rational number.\\

This means that it can be expressed as $\dfrac{a}{b}$ where $a\; and \;b$ \\are relatively prime integers (that means they have no common factors other than 1)\\

let $\sqrt5=\frac{a}{b}}$ Where a and b are relatively prime. \\

$\dfrac{a^2}{b^2}=5$\\\\

$a^2=5b^2$\\\\

Hence $a^2$ is a multiple of 5\\

Hence $a$ is a multiple of 5\\

so $a=5c$ for some integer c\\

$5b^2=a^2=(5c)^2=25c^2$\\

Dividing through by 5 this means\\

$b^2=5c^2$ \\

Hence $b^2$ is a multiple of 5\\

Hence $b$ is a multiple of 5\\\\$$

$$Now we have found that a and b must both be multiples of 5\\

This is a contradiction since we started with the premise that there existed an a and b such that a and b are relatively prime.\\

Therefore $\sqrt5$ cannot be a rational number.\\

Therefore $\sqrt5$ is an irrational number.$$

Melody Aug 31, 2014

#4**+5 **

Test this by assuming √5 is rational. If it is rational then it can be expressed in lowest terms as x/y.

$$\ Let \ (\frac{x}{y})^2 = 5 \\\

\hspace{20pt}\ x^2 / y^2 \ = 5 \\\

\hspace{20pt}\ x^2 = 5y^2 \\$$

Here the factor 5y^{2}, on the right, has an additional prime. One side of the equation has an odd number and the other has an even number. Composite numbers cannot have both an even and odd number of prime factors because all composite numbers have a unique prime factorization.

These two factors “compete” for balance into infinitesimal infinity.

This is an irrational number, a surd. As CPhill might say, something that cannot be heard or uttered.

~~D~~

DavidQD Aug 31, 2014

#6**+5 **

The equation is set up as attempt to balance, with a constant, the ratio of two hypothetical integers. The constant extends, in diminishing value, into infinity. In other words, no integers exist that define this ratio.

It is the middle ground of light and shadow.

~~D~~

DavidQD Aug 31, 2014

#7**+5 **

You are being too obtuse for me I am afraid David.

Here the factor 5y^{2}, on the right, has an additional prime. One side of the equation has an odd number and the other has an even number. Composite numbers cannot have both an even and odd number of prime factors because all composite numbers have a unique prime factorization.

$$x^2=5y^2$$

additional prime to what?

One side has an odd number do you mean of factors?

---------------------------------------------------

**Umm. Is this what you are saying?**

x^2 has an odd number of factors where as 5y^2 has an even number of factors.

This is impossible because they are the same number and must have the same number of factors

Therefore there are no integers x and y that meet this criterion.

Therefore sqrt5 is not rational

Therefore sqrt5 is irrational.

------------------------------------------------------------

Melody Aug 31, 2014

#8**+5 **

*Composite numbers cannot have both an even and odd number of prime factors because all composite numbers have a unique prime factorization.*

That is true. This solution assumes that it is, and proves it is not a composite number.

*This is impossible because they are the same number and must have the same number of factors.*

The “x” and “Y” values have the same number of factors, the equation does not. They cannot because the total factors are odd. Only at an infinite, diminishing point where the “constant” --the additional factor, converges does the equation balance. That value, in this case, is the SQR( 5).

I do not know which side has the even or odd, only that both exist. The total factors are odd, because adding all the factors, except for the 5 is the same as multiplying by 2, making the total an even amount, then add the one additional factor (the 5), to the count, makes it odd.

*Therefore there are no integers x and y that meet this criterion.*

That is essentially correct.

**This is conjecture**, but the mathematics that communicate this may be related to the maths that relate why the set of all integers is equal in number to the set of all even (or odd) integers.

~~D~~

DavidQD Aug 31, 2014

#9**0 **

**I do not know which side has the even or odd, only that both exist.**

Huh? I am afraid you talk in riddles David.

Any square number has to have an odd number of factors because all factors come in pairs but the squared one is doubled. Multiply a square number by 5 and it has to have an odd number plus one more. That is it has to have an even number of factors.

Isn't what I said right?

Melody Aug 31, 2014

#10**+8 **

Best Answer

Here's another proof:

Let's suppose that there are two positive, relatively-prime integers, call them a and b, such that a/b = √5 with a > b.

Multiply both sides by b, so we have

a = b√5

Now square both sides

a^{2} = 5b^{2}

And note that we can write 5b^{2} as 4b^{2} + b^{2 } Now, subtract b^{2} from both sides....so we have...

a^{2} - b^{2} = 4b^{2}

Now, note that the right side is even, and suppose that a and b are both even. But, this is impossible since they are relatively prime and thus, have no factors in common.

And neither can one be odd an the other even, because the left side would be odd, but the right side is even.

The only possibility left is that both are odd. But this can't be so for the following reason: Factor the left side as (a + b) (a - b). And if a and b are odd, then (a + b) is even and so is (a - b) - (remember that a > b, so a - b > 0).

So we have

(a + b) ( a - b) = 4b^{2} Now divide both sides by 4 ....... and we can write

[(a + b)/2] [(a - b)/2] = b^{2}

Now, if (a-b)/2 is odd, then (a+b)/2 is even ..... and vice-versa. To see this, let (a - b)/2 = 2k + 1. Then a - b = 2(2k + 1) .... Add b to both sides..... a = 2(2k + 1) + b....Now add b again to both sides and we have a + b = 2(2k +1) + 2b.......Now, divide by 2 on both sides, and we have (a+b)/2 = (2k + 1) + b, and both sides are even. Likewise, it can be proved that if (a +b)/ 2 is odd, then (a-b)/2 is even.

But, an even times an odd is always even...... but b^{2} is odd.

Thus......a and b aren't both even or both odd, and neither are they of different parity. and nothing else can be possible. Thus, our assumption that there exists two positive, relatively-prime integers such that a/b = √5 must be false. And thus, √5 must be irrational.

CPhill Aug 31, 2014

#12**0 **

Thanks Melody.

This was an attempt to demonstrate that because an integer squared (x^2) (where all such numbers have an odd number of positive divisors) is equal to another number squared (y^2), plus an additional factor (5) that the value of (x) and/or (y) cannot be integers. This means it is equal to a number that has an even number of divisors, or it does not have an even number of divisors to begin with. Because of this contradiction, the number cannot be an integer. Extending on this was an attempt to explain why an irrational number had characteristics of both even and odd numbers.

The explanation I presented is incomplete. This has been rolling around in my head for years, and it seemed functional, until an examination in a proper light. On closer inspection, it seems it is an incomplete amalgamation of proofs by contradiction and a hint of a proof by infinite decent thrown in for good measure (Fermat would be appalled). Mostly, it is a succession of incongruous comparisons that are not made compatible mathematically. Very lame. This is worse than mixing up metaphors. So, thank you Melody, I now see the darkness at the end of the tunnel.

Your proof is to the point clear. CPhill’s, detailed and elegant, proof clearly demonstrates the paradox of (even and odd) parity.

Somewhere, there is a direct proof (instead of proofs by contradiction) of irrational numbers. Probably the best place to look is in Andrew John Wiles’ proof of Fermat’s Last Theorem. These maths are far outside of my skill set, and always will be. Even so, after dozens of readings, I have gleaned insights from them.

To find a direct proof for irrational numbers could lead to maths that close the gaps on Cantor’s Alephs of infinites. Considering there are more irrational numbers than integers, the value of this proof may be enormous. Adding to this, Wiles’ proof includes proofs of the Modularity Theorem for semistable elliptic curves, with these kinds of proofs, all three of Georg Cantor’s infinites may converge. Wiles’ proof seems like an excellent place to start.

Noting the post on Fermat, there is one question that remains about Fermat’s Last Theorem: Did Fermat really have the remarkable proof he wrote of in the margin of his book? Considering it took advanced, late twentieth century mathematics to solve it, I seriously doubt it. However, it is an intriguing thought that seventeenth century math may hold a solution.

This forum is great. The mathematicians who visit and especially the ones who live here are truly extraordinary.

Thank you Melody and CPhill!

~~D~~

DavidQD Sep 2, 2014

#13**0 **

DavidQD

From my brief knowledge about the proof of Fermat's Last Theorem....I don't think that most math histrorians believe that Fermat had enough math available to him in his era to prove his own conjecture. I believe - if I remember correctly - that Wiles brought two seemingly unrelated areas of math together (ellliptical curves and something else??) to establish a proof.

I once asked one of my math professors if he could explain Wiles' proof.....he told me that it was WAY above his pay grade.....I knew then that it must be pretty difficult to understand !!!!

CPhill Sep 2, 2014