Prove that √2+√3 is irrational?
Assume it's rational. Then there are integers m,n such that m/n = √2 + √3 Square both sides: m^2 / n^2 = 5 + 2√6 "Solve" for √6: √6 = (m^2 - 5n^2) / (2n^2) so if √2 + √3 is rational, then so is √6. Let a and b be the integers with gcd(a,b) = 1 such that a/b = √6 Square both sides and multiply by b^2: a^2 = 6b^2 Now, the right side is divisible by 2, so a^2 is divisible by 2, which then implies that a is divisible by 2 (since 2 is prime). Therefore we can write a=2k for some integer k: 4k^2 = (2k)^2 = 6b^2 Divide by 2: 2k^2 = 3b^2 Now the left side is divisible by 2, so 3b^2 is divisible by 2, from which it follows that b is divisible by 2. However, this would mean that 2 divides gcd(a,b) = 1. Contradiction. ------ Therefore √6 is irrational, and therefore √2 + √3 is irrational also.
Prove 2^(1/4) is irrational and that 2^(1/3) is irrational and also that cos(pi/8) & cos(1) are irrational?
rational means that a number can be written in the form of a/b such that gcd (a,b) = 1. irrational means it cannot be written in such form of a/b.... it has a long continious decimal place which cannot be converted to the form a/b example: sqrt(2) is irrational sqrt(4) is rational because sqrt(4) =2 = 2/1 and gcd (2,1) = 1
Can you prove that 3÷2√3 is irrational?
3/2√3=√9/√12=√(9/12)=√(3/4)=√3/2√3/2 is irrational therefore 3/2√3 is irrationalThus proven,
Prove that log 5 at base 3 is irrational?
Proof by contradiction:Let log 5 at base 3 is rational, say a/b wherea and b are positive integers(check that a/b has to be positive).So[math]3^{a/b} = 5[/math][math]3^a = 5^b[/math]Therefore 3 must divide 5.But 3 and 5 are co-prime.So, our assumption is wrong.Hence it is irrational.
Prove that 1+3sqrt2 is irrational?
This presupposes that √2 is irrational. Suppose that 1 + 3√2 is rational. You could then say that for integers a and b, b being non-zero, you have: 1 + 3√2 = a/b 3√2 = a/b - 1 = (a - b)/b √2 = (a - b)/(3b) That is, we deduced that √2 is rational, which is a contradiction. Thus, the number must be irrational.
How would I prove that 2 + sqrt 3 is irrational?
Suppose that 2 + sqrt 3 is a rational number. Since 2 is rational, sqrt 3 must be rational since the difference of two rational numbers is rational. So sqrt 3 can be expressed as m/n, where m,n are integers with gcd(m,n)=1. Taking squares of both sides, we get m^2 / n^2 = 3. Thus, m^2 = 3*(n^2) (***). Since gcd(m,n)=1, 3 divides m^2 and so 3 divides m. Now write m=3k, k is an integer and gcd(k,n)=1, of course.Then (***) becomes 9*k^2 = 3*n^2 => 3*k^2 = n^2. Since gcd(k,n)=1, we conclude that 3 divides n^2,and so n. Then both m and n are divisible by 3, but we know that gcd(m,n)=1. Contradiction.
How can I prove that [math]\sqrt{3} + 2[/math] is irrational?
[math]\text{Lemma }1[/math]:If [math]a[/math] is a rational number and [math]x[/math] is an irrational number,then [math]a+x[/math] is also an irrational number.[math]\text{Proof}[/math]:Suppose for the sake of contradiction that [math]a[/math] is rational and [math]x[/math] is irrational but [math]a+x[/math] is rational.Since [math]a+x[/math] is rational there exists integers [math]p[/math] and [math]q \neq 0[/math] such that [math]a+x=\dfrac{p}{q}[/math].Subtracting [math]a[/math] from both the sides,we get [math]x=\dfrac{p}{q}-a[/math].Because [math]a[/math] is rational there exists integers [math]m[/math] and [math]n \neq 0[/math] such that [math]a=\dfrac{m}{n}[/math] and hence [math]x=\dfrac{p}{q}-\dfrac{m}{n}=\dfrac{np-mq}{nq}[/math].Since [math]m,n,p,q[/math] are integers,both [math]np-mq[/math] and [math]nq[/math] are also integers and [math]x[/math] is written as ratio of two integers.Thus [math]x[/math] is a rational,which is a contradiction to the fact that [math]x[/math] is irrational.Hence [math]a+x[/math] is irrational.[math]\text{Lemma }2[/math]: [math]\sqrt{3}[/math] is irrational.User has already given a proof for this Lemma. You can see User's answer to How can I prove that [math]\sqrt{3} + 2[/math] is irrational?[math]\text {Claim }[/math]: [math]2+\sqrt{3}[/math] is irrational.[math]\text{Proof}[/math]:Use Lemma 1 with [math]a=2[/math] and [math]x=\sqrt{3}[/math].
How do I prove that “arccos (1/√3) /π” is irrational?
As with most proofs of irrationality, we use a technique called Proof by Contradiction. Suppose, to the contrary, that there exist integers [math]p,q[/math] with [math]q \ge 1[/math] such that [math]\dfrac{\arccos(1/\sqrt{3})}{\pi} = \dfrac{p}{q}[/math].We can rearrange this equation slightly: [math]2q\arccos\left(\dfrac{1}{\sqrt{3}}\right) = 2p\pi[/math]. Therefore,[math]\cos\left(2q\arccos\left(\dfrac{1}{\sqrt{3}}\right)\right) = \cos(2p\pi) = 1[/math]Now define [math]T_n(x) = \cos(n\arccos(x))[/math] for [math]n \in \mathbb{N}[/math]. This is known as a Chebyshev Polynomial of the First Kind. In case you’re not familiar with the concept, you can read about them here: Chebyshev polynomials - WikipediaFor this proof, we only need to know two properties of these polynomials:[math]T_n(x)[/math] is a polynomial in [math]x[/math] with integer coefficients;The leading coefficient of [math]T_n(x)[/math] is [math]2^{n-1}[/math] for [math]n \ge 1[/math].Let [math]f(x) = T_{2q}(x) - 1[/math]. Since [math]T_{2q}(x)[/math] has only integer coefficients, so does [math]f(x)[/math]. From the last equation, we know that [math]f(x)[/math] has a root at [math]x = \dfrac{1}{\sqrt{3}}[/math]. The minimal polynomial of this number is [math]3x^2 - 1[/math]. Therefore, by properties of minimal polynomials, [math]3x^2 - 1[/math] is a factor of [math]f(x)[/math]. Therefore, by Gauss’ Lemma, the leading coefficient of [math]f[/math] must be divisible by [math]3[/math]. However the leading coefficient of [math]f[/math] is the same as the leading coefficient of [math]T_{2q}[/math], which is [math]2^{2q-1}[/math]. Thus [math]2^{2q-1}[/math] is divisible by [math]3[/math], which is impossible.Since we reached a contradiction, our original assumption was wrong. Therefore [math]\dfrac{\arccos(1/\sqrt{3})}{\pi}[/math] is irrational.QED
How to prove 7^(1/3) is an irrational number?
Simple assume[math]7^{\frac13}[/math] to be a rational number, this means that you can write it as an integer fraction [math]\frac{p}{q}[/math]. Since every fraction can be simplified (dividing out the comment factor, making [math]p[/math] and [math]q[/math] co-prime), let us assume that [math]\frac{p}{q}[/math] is simplified.Ok[math]7^{\frac13}=\frac{p}{q} \mid t^3[/math][math]7=\frac{p^3}{q^3} \mid \times q^3[/math][math]7q^3=p^3 \implies 7 | p^3 \implies 7 | p[/math]So with a little algebra we have shown [math]p^3[/math] is a multiple of [math]7[/math], since [math]7[/math] is a prime number, [math]p[/math] itself must be a multiple of [math]7[/math].But now [math]p=7r \implies p^3=343r^3[/math]so we have[math]7q^3=343r^3 \mid \div 7[/math][math]q^3=49r^3[/math]But wait [math]49=7^2 \therefore \implies 7 | q[/math]So we have a fully simplified fraction [math]\frac{p}{q}[/math] but both [math]p[/math] and [math]q[/math] are divisible by [math]7[/math]. This is a contradiction, therefore our assumption that [math]7^{\frac13}[/math] is a rational number is invalid, hence [math]7^{\frac13}[/math] is irrational.Does this proof seem familiar? It should it is the famous proof that [math]\sqrt{2}[/math] is irrational, just a little bit rewritten, after all [math]7^{\frac13}=\sqrt[3]{7}[/math].I have an even more general proof where I show that [math]\forall m, n \in \mathbb N^1 \left (\sqrt[m]{n} \not \in \mathbb N \implies \sqrt[m]{n} \not \in \mathbb Q \right )[/math] (for all positive integer [math]m[/math] and [math]n[/math] the [math]m[/math]-root of [math]n \ \left (\sqrt[m]{n} \right )[/math] is either a natural number or irrational) in an old answer of mine.And given the identity:[math]\sqrt[c]{a^b}=\left (\sqrt[c]{a} \right )^b=a^{\frac{b}{c}}[/math]and[math]\left (\frac{a}{b} \right )^c=\frac{a^c}{b^c}=\frac{a^c \times b^{1-c}}{b}[/math]and:[math]a^{-b}=\frac{1}{a^b}[/math]you should be able to show the irrationality/rationality of every number of the form [math]r^q \mid q, r \in \mathbb Q \land r \gt 0[/math].