Question

2root3 - 7root2 is irrational prove that

Answer 1

2 answers · Mathematics 

 Best Answer

It's the difference of two irrational numbers, so... Just kidding. :-) 

Let x = 2sqrt(3) - 7sqrt(2). Then: 

x^2 = 12 - 28sqrt(6) + 98 
x^2 = 110 - 28sqrt(6) 
x^2 - 110 = -28sqrt(6) 
(x^2 - 110)^2 = 4704 
x^4 - 220x^2 + 12100 = 4704 
x^4 - 220x^2 + 7396 = 0 

So, we have a polynomial of which x is a root. Essentially we want to use the rational roots theorem. It says that, given a polynomial of integer coefficients, the rational roots are of the form p/q, where p is a factor of the constant term (in this case 7396) and q is a factor of the leading term (in this case 1, so either 1 or -1). So, if x is a rational root, it must be an integer factor of 7396. 

What are the integer factors of 7396? It doesn't matter. The fact that x is an integer gives us enough. If we use a calculator, we can approximate 2sqrt(3) - 7sqrt(2) to put it between -7 and -6. Since calculators are not rigorous proofs, we can prove that: 

-7 < 2sqrt(3) - 7sqrt(2) < -6 
<===> 6 < 7sqrt(2) - 2sqrt(3) < 7 
<===> 36 < 110 - 28sqrt(6) < 49 
<===> 28sqrt(6) < 74 and 28sqrt(6) > 61 
<===> 61 < 28sqrt(6) < 74 
<===> 61^2 < 28^2 * 6 < 74^2 
<===> 3721 < 4704 < 5476 

which is clearly true. Therefore, the number lies strictly between the integers, which contradicts x being rational. 

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If you don't want to assume the rational roots theorem, it's easy enough to prove. Suppose x = p/q for integers p, q. Then we can assume that gcd(p, q) = 1, i.e. it's in lowest terms. Then:

(p/q)^4 - 220(p/q)^2 + 7396 = 0 
p^4 - 220p^2q^2 + 7396q^4 = 0 
p^4 = 220p^2q^2 - 7396q^4 
p^4 = (220p^2 - 7396q^2)q^2 

Thus q divides p^4. Therefore, because gcd(p, q) = 1, q must divide one of the factors of p^4, so q must divide p. But then q is a factor of both p and q, the only ones of which are 1 and -1. Thus, x is an integer. 

Alternatively, if x is rational, x^2 is also rational, and is a root of the polynomial: 

y^2 - 220y + 7396 


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Answer 2

see the answer in the picture