Proving Algebraic Nature of Root 3 + Root 2

[lokisapocalypse]

I need to show that this is an algebraic number.

In other words,

I need to show: an*x^n + an1*x^(n-1) + ... + a1 * x^1 + a0 * x^0 =
where the a terms are not ALL 0 but some of them can be.

Like for root 2 by itself,

I have 1 * (root 2) ^ 2 + 0 * (root 2)^1 + -2 * (root 2) ^ 0
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[robert Ihnot]

Well, I take it you mean to write an equation having the root $$\sqrt2 +\sqrt3$$

See the homework section: https://www.physicsforums.com/showthread.php?t=62232

 

[M98Ranger]

To prove the algebraic nature of root 3 + root 2, we need to show that it can be expressed as a polynomial with rational coefficients.

First, let's simplify root 3 + root 2 by using the distributive property:

root 3 + root 2 = (1 * root 3) + (1 * root 2)

Next, we can use the property of radicals that states the product of two radicals is equal to the radical of the product of the two numbers inside the radicals.

So, (1 * root 3) + (1 * root 2) = root (3 * 2)

Simplifying further, we get root 6.

Now, we can express root 6 as a polynomial with rational coefficients:

root 6 = 0 * x^3 + 0 * x^2 + 0 * x + 1 * x^0

This shows that root 6 is an algebraic number, and since root 3 + root 2 can be simplified to root 6, it is also an algebraic number.

Therefore, we have proven that root 3 + root 2 is an algebraic number, as it can be expressed as a polynomial with rational coefficients.