Yes , an equation which combines i , pi , e , 1 and 0 !  Very beautiful equation .

Hmmm I kinda fathom the question, but I have no idea on how to show that it's rational or not. A very tricky one I'd say....

Can you name that beautiful equation?
Hint: See Post #226 on Page 12.

Try a simpler one then : Proof , or disproof that the product of two irrational numbers results in a rational number .

Yes , an equation which combines i , pi , e , 1 and 0 !  Very beautiful equation .

I would appreciate it if you could provide tutorials on Logic .

https://fbcdn-sphotos-h-a.akamaihd.net/hpho...650586348_n.jpg

The link will bring you to a page of 2013 Further Mathematics STPM Repeat Paper . In Q1 , it asks for the validity of propositions , but in my book , it says that a proposition and an argument are two different things . A proposition can be true or false , valid and invalid are incorrect terms to refer to propositions .

I know how to do (a) and (b) , but what about the conclusion ?

Can you name that beautiful equation?
Hint: See Post #226 on Page 12.

Hmmm I kinda fathom the question, but I have no idea on how to show that it's rational or not. A very tricky one I'd say....

Wait , I think I see something .

Suppose that (√2)^(√2) is irrational .

Let (√2)^(√2) = a , where a is an irrational number .

Raising the powers of both sides by √2 , we have (√2)^2 = a^(√2) ;

a^(√2) = 2  ( a is irrational , √2 is also irrational but 2 is rational ) .



Call me a nerd , but this is much more exciting than a lot of things I've seen !

The solution makes use of the Gelfond-Schneider theorem.
Correct, but what happens when a≠b?

The solution makes use of the Gelfond-Schneider theorem.
Correct, but what happens when a≠b?

I'm not so sure about this, but I might give it a try. We know that √2 is irrational. But if we multiply √2 with √2:
√2 x √2 = 2/1
Hence, 2 in this case, is rational.

The conclusion is that for every x∈R, ∃ an inverse y∈R ∋x+y∈R.

Didn't think of that . 

Is there a general proof ?

∋ means 'such that' ?

Any reasoning behind this , maybe a more detailed explanation ? I'm sorry , I've a lot to go .

The solution makes use of the Gelfond-Schneider theorem.

It's not necessary to provide general proof. Existence is shown if you can provide just one example where the property holds.
Sorry, creature of habit. ∋ means such that. For any real number, an additive inverse exists within the set so that that number added to it's inverse will yield the additive identity, 0. This is a basic property of the vector space R of real numbers. You can find more info on this in books on linear algebra.

What about a general proof ?

Can I say :

For every x∈R, ∃ an inverse y∈R such that x+y = 0 instead of x+y∈R ?

Wait, did I miss something?   If I were to multiply an irrational number(√2) with itself, then I'll get a rational number, which is 2. So, the product of two irrational numbers is rational. Did I left out anything vital?

Wait, did I miss something?   If I were to multiply an irrational number(√2) with itself, then I'll get a rational number, which is 2. So, the product of two irrational numbers is rational. Did I left out anything vital?

Your concern for the question is this - is the irrational power of an irrational still irrational?

Try this out :

√2*√3 , do you still get a rational number ?

An irrational to an irrational power may be rational, as what has been shown previously by maximR,(which also involves multiplying the Gelfond–Schneider constant with √2)

But, if in the case of √2*√3, I don't think the result is rational. Correct me if I'm wrong though.

The product of irrationals aren't always rational. As with the original question, one only need show an example to prove this statement true.

For every x and y such that both are irrational , their product is a rational number .

If the statement is revised as above , can one example prove the statement ?