Can we construct three irrational numbers a,b,ca,b,c such that a+b+c∈Qa+b+c \in \mathbb Q?

This is rather easily shown to be possible if no constraint is put on a,b,c. However, is it also possible under the following constraint: a,b and c can not be rational multiples of each other. If that constraint is too tight to allow such a construction, would it be possible if we loosen it a … Read more

A simple proof for non-rationality of number n√1+qn\sqrt[n]{1+q^n}

I’m going to prove the following fact (preferably with an elementary simple method): Let n>2 be a natural number. Then for all q∈Q+∖{0} we have n√1+qn∉Q+∖{0}. Assuming the contrary, we then have (1+anbn)=cndn, where q=ab and n√1+qn=cd. Now multiplying both sides of the above equation by bndn we get a contradiction according to Fermat Last … Read more

Prove that 1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}} is irrational

I tried to follow this example to resolve this exercise, this example to resolve this exercise, however if: 1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=x,then 3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=(x-1), and (3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}})^5=3720+2985\cdot 3^{\frac{1}{5}}+2385\cdot3^{\frac{2}{5}}+1905\cdot3^{\frac{3}{5}}+1545\cdot3^{\frac{4}{5}} but i cannot factorize the same way since that would result like 1545x+2175+1350\cdot 3^{\frac{1}{5}}+840\cdot3^{\frac{2}{5}}+360\cdot3^{\frac{3}{5}}=(x-1)^5 Answer Observe that 3^{1/5} x = 3^{1/5} + 3^{2/5} + 3^{3/5} + 3^{4/5} + 3 = x + … Read more

Is x^yx^y always irrational if xx is rational and yy is irrational?

Prove or disprove: “If x is a rational number, and y is an irrational number then x^y is irrational” I am stuck with this, these are my steps. let x=2 and y=\sqrt{2} \implies x^y = 2^{\sqrt{2}} now if x^y = 2^{\sqrt{2}} is irrational then we are done. But if this is rational then we can … Read more

Prove that $(\sqrt3+2)^m$ is not a natural number for all natural numbers $m≥1$

The aim of this question is to show this lemma: Prove that $(√3+2)^{m}$ is not a natural number for all natural numbers $m≥1$. Answer Because $(\sqrt{3}+2)^{m}+(-\sqrt{3}+2)^{m}$ is an integer and $0<(-\sqrt{3}+2)^{m}<1$. AttributionSource : Link , Question Author : DER , Answer Author : Chen Jiang

Proving the irrationality of √5\sqrt{5}: if 55 divides x2x^2, then 55 divides xx

I am working on proving that √5 is irrational. I think I have the proof down, there is just one part I am stuck on. How do I prove that x2 is divisible by 5 then x is also divisible by 5? Right now I have 5y2=x2 I am doing a proof by contradiction, where … Read more

Is $\frac{\sqrt7}{\sqrt[3]{15}}$ rational or irrational?

Is $\frac{\sqrt7}{\sqrt[3]{15}}$ rational or irrational? Prove it. I am having a hard time with this question. So far what I did was say, assume it’s rational, then $$\frac{\sqrt7}{\sqrt[3]{15}}=\frac{x}{y} \Rightarrow \sqrt{7}y=x\sqrt[3]{15}$$ I then showed the product of a rational number and an irrational number is irrational so the expression above is irrational on the left and … Read more

Prove that if nn is a positive integer then √n+√2\sqrt{n}+ \sqrt{2} is irrational

Prove that if n is a positive integer then √n+√2 is irrational. The sum of a rational and irrational number is always irrational, that much I know – thus, if n is a perfect square, we are finished. However, is it not possible that the sum of two irrational numbers be rational? If not, how … Read more