How to prove "Square root of 3" as irrational number?

Question

I need different methods of proving "Square root of 3" as irrational number.

Answer 1

We prove that by contradiction.
This means, we suppose √3 is rational and then get to something that contradicts this assumption.

If √3 is rational, we can express it as a/b, a fraction in its lowest terms (i.e., gcd(a,b) = 1). Also, b is not 1, because that would mean √3 is an integer, which it obviously is not.

Now square both sides.
√3 = a/b
3 = a²/b²

since gcd(a,b) = 1, then gcd(a²,b²) = 1 (has to be).

But from 3 = a²/b² we can express a² = 3b².

So a²/b² = 3b²/b² -> now this fraction is obviously NOT in its lowest terms (remember, b is not 1) -> hence the contradiction.

=> √3 is not rational. It is irrational.

Hope this helps.

Answer 2

pointy has the right idea, but you still have to prove that \sqrt{3} is irrational. You can prove that the square root of any prime number is irrational in a similar way that you prove it for the square root of 2. Suppose that (m/n)^2 = p , where p is a prime. (You will use p=3). Further, suppose that the fraction (m/n) is fully reduced : GCD(m,n)=1 . Then: m^2 = p*n^2 .Thus, p divides m^2. Since p is prime, this means that p divides m. Say m=k*p. Now we have the equation: (k^2)(p^2)/(n^2) = p . Which gives us: p*k^2 = n^2. For the same reason as before, we get that p divides n. Say j*p=n . Our original fraction: (m)/(n) = (k*p)/(j*p) . Our original assumption that the fraction was in most simplified form has been contradicted. Thus, the square root of p must be irrational.

Answer 3

Irrational number depicts "endless nature of a prime number (without involving any recurring groups)". Take "3" as 2.999999999999999999999... means any higher number of nines after decimal point!

We can also take it as 3.0000000...........0000000001

Square root of both said numbers are irrational!

2.9999999999999999999999999999999999999999...is 3
1. 7 3 2 0 5 0 8 0 7 5 6 8 8 7 7 2 9 3 5 2...sq.rt

17^2= 289...+10
173^2=29929...+70
1732^2=2999824...+175
17320^2=299982400...+17599
173205^2=29999972025...+27974
17320508^2=299999997378064...
+2621935

1) You will find "a consistent increase of remainder digits when "digits in square root" increases! There is no chance of reversing said trent (whatsoever) and so said square roots are irrational!

Look at square of "a-unit more than17320508"----> 17320509^2 = 300000032019081...(-320190080). There are appreciable variation from 3.00000000000001.

It implies, either by a relating to (a infinitesimal "minus or plus"- value change from base 3,0000...), true square root digits relate appreciably away numbers from mean 3.00, or 3.0000, or 3.000000... it implies "a nature of prime number 3" is followed by "square root 3", which is irrational!

Answer 4

1) the square root of three has a non terminating continued fraction expansion,rational numbers do not.

2)If one number was both an interger square and three times an integer square it would have both an even and an odd number of prime factors which is a contradiction.

3)If three was the quotient of two relatively prime squares then each would be divisible by 3, a contradiction.

4)If one square was three times another then smaller squares could be found with this property and there would be no smallest squares with this property, a contradiction.

Answer 5

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How to prove "Square root of 3" as irrational number?
I need different methods of proving "Square root of 3" as irrational number.

Answer 6

leave it as

(square root)3