Proof: √3 + √2 is irrational

[mr.schloopka]

The second proof is easy.
sqrt(2)+sqrt(3)=1.41 + 1.73 = 3.14 = pi
And everyone knows pi is irrational.

[ELLA-owtm]

You make math feel so much more accessible to me, thank you for your amazing videos!

[JohnZwiers]

I love all the Algebra 1 connections in your proof. It is a great way to show the beginning algebra tools are quite valuable

[sachinphile]

funny thing, i was just proving this and then this got recommended !

[rohitchaoji]

When you were looking for a number aside from 10, I said to myself "21!", and then you said it too. Great minds think alike! Or, mediocre minds something think like great minds and it helps them boost their confidence for two minutes of the day.

[brandonbutler]

I loved my college professor who taught Proofs and Analysis, but this is so good! I finally understand this proof!

[David__U]

Mistake on the second to last line of the second proof: you wrote that the expression is Not an element of the irrationals, which is the same as saying it IS rational, which is not a contradiction. You probably didn't want the tilda.

[dougowner]

Awesome, as always! By the way, Eddie, which model is that Tablet you use for the classes? It looks wonderful.

[NickKravitz]

In Unix context the tilde means home folder. And my home folder contains lots of irrational items.

[Kindiakan]

Might be clearer to show that since Q is closed under the four arithmetical operations, sqrt(6) = ((sqrt(3) + sqrt(2))^2 - 5)/2 is rational, contradicting the first part.

[vishalmishra]

*General Theorem* - Every non-negative integer is either a perfect square (whose roots are +/- an integer) or a square of an [+/-] irrational number. i.e. for any co-prime integers p and q, if (p/q)^2 = n = some non-negative integer then q = 1 (the rational number has to be an integer). √6 (and more generally any linear combination of square-roots of non-perfect square integers) can be proved to be irrational using this *General Theorem* . No point in proving any specific case such as √6 or √3 +/- √2.

[abhinavdiddigam]

After 6=p^2/q^2
We can also transpose q^2 to the left and can say that 6 is a factor of P^2 hence a factor of p too. As 6 is a factor, p=6c for some integer c
Substituting that in the original equation
(6c)^2= 6q^2
After cancellations, we get that 6 is a factor of q^2 and hence a factor of 2 as well.
But this contradicts with the fact that.p and q are co-primes

[kindiakmath]

For (b) I think we can make the argument airtight by writing sqrt(6) = ((sqrt(3)+sqrt(2))^2 - 5)/2. Under our supposition, the RHS is rational, but by (a), LHS is irrational, which gives an explicit contradiction :)

[sazgar]

You had a misprinted sir, the square of the sum is not belong to rational. And you wrote irrational instead of rational. Before last line. By the way, you are the best.

[AchtungBaby]

There's a much easier & faster method of proof to show that the square root of ANY non-square number is irrational.
The square root of a positive integer n is the solution of the polynomial x^2 - n = 0.
By the Rational Root theorem, the only rational candidates are factors of n, and since none of them work, the solution must be irrational! QED :)

[aashsyed]

5:25 we can do that because if p and q share factors, then we can cancel them such that they are in the lowest form. but that is the same as if they share no factors. so that is why we can assume that statement.

[SHADOWFACTSINTELUGU]

I am in 10th. And I feel more comfortable to apply our method to prove √6 is irrational.. 😂 I would solve this in 2min... But I got confused by seeing all these 😂😂

Thank you for a great explanation of mathematical logic.

[Curufin]

In my opinion, instead of doing all those standard proofs about sqrt(2) (or in this case sqrt(6)) proofs it is much easer seeing that the n-th root of an integer is either an integer or irrational. This requires no calculations and is more insightful.

[RexxSchneider]

Where is the *proof* that 5 + 2*sqrt(6) is irrational? We've proved that sqrt(6) is irrational, but nothing you've shown has proven that a linear combination of rationals with a single irrational is irrational.
That is true - and not difficult to prove - but you can't just say "because it has an irrational tucked away inside" the expression. Rigorous proofs don't rely on "I say so".