How can i prove that $\sqrt3/\sqrt2$ is irrational?
Thank you for your help!
Asked
Active
Viewed 192 times
0
-
2Do you know how to prove that $\sqrt{2}$ is irrational? – Michael Biro Nov 20 '16 at 16:55
-
Look here. – Tacet Nov 20 '16 at 16:56
-
Yes i know how to prove this. – Schubertliszt Nov 20 '16 at 16:57
-
1Otherwise $3n^2 = 2m^2$, $m,n$ coprime, so that $n$ is even. But then $12p^2 = 2m^2$, or $6p^2 = m^2$ implies $m$ is even, so they're not coprime, contradiction – MathematicsStudent1122 Nov 20 '16 at 16:59
3 Answers
4
If $\sqrt{3}/\sqrt2=m/n$ ($m$ and $n$ positive integers) then $$ 3n^2=2m^2. $$ But that is impossible, because the prime decomposition of the rhs contains an odd number of $2$, while on the lhs there is an even number of $2$.
Intelligenti pauca
- 50,470
- 4
- 42
- 77
-
-
1In the prime decomposition of a square number, all prime factors are repeated an even number of times (including 0 times, of course). But on the right hand side there is an additional factor of 2, which is not present on the left hand side. – Intelligenti pauca Mar 28 '17 at 21:05
-
Thank you but please what does mean odd number of 2 and even number of 2 is that mean : (2k) and (2k+1) but i cant see "of 2" meaning would you clarify that ? – Educ Mar 28 '17 at 21:09
-
1$2\cdot2\cdot2$ is a product with an odd number of 2's, $2\cdot2\cdot2\cdot2$ is a product with an even number of 2's. – Intelligenti pauca Mar 28 '17 at 21:18
4
If $\frac{\sqrt3}{\sqrt2}$ is rational then so is its double, $\sqrt6$.
$\sqrt6$ is the root of $x^2-6$, and by the Rational Root Theorem the only possible rational roots of this polynomial are $\left\{\pm1,\pm2,\pm3,\pm6\right\}$, none of which are roots.
robjohn
- 345,667
-
The "doubling" is equivalent to applying the AC-method to $,2x^2 -3,$ to obtain the monic $,X^2 -6,,$ for $,X = 2x.\ $ – Bill Dubuque Nov 22 '16 at 23:42
-
Of course, we can simply apply the Rational Root Theorem to $2x^2-3$ and get $\left{\pm\frac12,\pm1,\pm\frac32,\pm3\right}$. However, it seemed so much easier to verify that the the integers in the list did not satisfy $x^2-6=0$. – robjohn Nov 23 '16 at 00:38
-
Right, but my point was to show readers how your idea of scaling to integral roots generalizes to arbitrary polynomials (a simple yet frequently useful method deserves to be better known). – Bill Dubuque Nov 23 '16 at 01:19
0
We prove by method of contradiction. Suppose the given no sqrt(3)/sqrt(2) is rational. Try rationalising the denominator sqrt(2) by sqrt(2). Then we get the no to be of the form sqrt(6)/2 which is rational by our assumption. Thus 2×(sqrt(6))/2 =sqrt(6) is also rational as product of two rational is again rational i.e sqrt(6) is rational which is a contradiction. This is due to our assumption that sqrt(3(/sqrt(2) is rational. Hence our assumption is wrong and so it is irrational
