There is a classic example here. Consider \(\displaystyle{A}=\sqrt{{2}}^{{\sqrt{{2}}}}\). Then \(\displaystyle{A}\) is either rational or irrational. If it is irrational, then we have \(\displaystyle{A}^{{\sqrt{{2}}}}=\sqrt{{2}}^{{{2}}}={2}\)

Juan Spiller

Answered 2022-01-08
Author has **5192** answers

asked 2022-01-06

\(\sqrt[31]{12}+\sqrt[12]{31}\) is irrational?

asked 2021-09-15

Prove or disprove that the product of two irrational numbers is irrational.

asked 2020-12-24

True or False?

1) Let x and y real numbers. If \(\displaystyle{x}^{{2}}-{5}{x}={y}^{{2}}-{5}{y}\) and \(\displaystyle{x}\ne{y}\), then x+y is five.

2) The real number pi can be expressed as a repeating decimal.

3) If an irrational number is divided by a nonzero integer the result is irrational.

1) Let x and y real numbers. If \(\displaystyle{x}^{{2}}-{5}{x}={y}^{{2}}-{5}{y}\) and \(\displaystyle{x}\ne{y}\), then x+y is five.

2) The real number pi can be expressed as a repeating decimal.

3) If an irrational number is divided by a nonzero integer the result is irrational.

asked 2021-12-03

Give an example of each of the following: (a) A natural number (b) An integer that is not a natural number (c) A rational number that is not an integer (d) An irrational number

asked 2022-01-04

How can I prove that the sum \(\sqrt{2}+\sqrt[3]{3}\) is an irrational number ??

asked 2022-01-05

Expansion of \(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}\)

I was asked to show that \(\displaystyle\forall{n}\in{\mathbb{{N}}}\) there exist a \(\displaystyle{p}\in{\mathbb{{N}}}^{\cdot}\) such that

\(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}=\sqrt{{p}}+\sqrt{{{p}-{1}}}\)

I used induction but it wasn't fruitful, so I tried to use the binomial expansion of \(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}\) but it seems I lack some insight to go further.

Any hint is welcomed.

I was asked to show that \(\displaystyle\forall{n}\in{\mathbb{{N}}}\) there exist a \(\displaystyle{p}\in{\mathbb{{N}}}^{\cdot}\) such that

\(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}=\sqrt{{p}}+\sqrt{{{p}-{1}}}\)

I used induction but it wasn't fruitful, so I tried to use the binomial expansion of \(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}\) but it seems I lack some insight to go further.

Any hint is welcomed.

asked 2021-12-11

I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if x is divisible by 3, then \(\displaystyle{x}^{{2}}\) is divisible by 3. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of 6?