The answer requires a bit of mathematics, but goes like this:The product of any 2 rational numbers is a rational number.The product of any 2 irrational number is an irrational number.The product of a rational and an irrational number is an irrational number!Therefore simple logic tells us that there are more irrational numbers than rational numbers. There is a way to structure this mathematically, and I believe it is called an "Inductive Proof".Interesting !I'm going to say "No".I reason thusly:-- For every rational number 'N', you can multiply or divide it by 'e', add it to 'e',or subtract it from 'e', and the result is irrational.-- You can multiply or divide it by (pi), add it to (pi), or subtract it from (pi),and the result is irrational.-- You can take its square root, and more times than not, its square root is irrational.There may be others that didn't occur to me just now. But even if there aren't,here are a bunch of irrational numbers that you can make from every rational one.This leads me to believe that there are more irrational numbers than rational ones.-------------------------------------------------------------------------------------------------------There are infinitely many more irrationals than rationals; this was proved by G. Cantor (born 1845, died 1918). His proof is basically:The rational numbers can be listed by assigning to each of the counting numbers (1, 2, 3,...) one of the rational numbers in such a way that every rational number is assigned to at least one counting number;If it is assumed that every irrational number can be assigned to at least one counting numbers (like the rationals), then with such a list it is possible to find an irrational number that is not on the list; so is it not possible as there are more irrationals than there are counting numbers, which has shown to be the same size as the rational numbers, thus showing that there are more irrationals than rationals.
an irrational number is any real number that cannot be expressed as a ratio a/b, where a and bare integers, with b nonzero, and is therefore not a rational number.Informally, this means that an irrational number cannot be represented as a simple fraction. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals. As a consequence of Cantor's proof that the real numbers are uncountable (and the rationals countable) it follows that almost all real numbers are irrational.[1]When the ratio of lengths of two line segments is irrational, the line segments are also described as being incommensurable By Paul Philip S. Panis
There cannot be a proof since your assertion is not necessarily true. sqrt(2)*sqrt(3) = sqrt(6). All three are irrational numbers.
There are uncountably infinite irrational numbers and their existence has been known for over 2500 years. In some cases, although their irrationality was suspected, rigorous mathematical proof took longer. Some notable events:sqrt(2): Pythagoreans (6th Century BCE).pi: Johann Heinrich Lambert (18th Century).e: Leonhard Euler (18th Century).In the late 19th Century, Georg Cantor proved that the number of irrationals is an order of infinity greater than the number of rationals.
sqrt(2) is irrational. 3 is rational. The product of an irrational and a non-zero rational is irrational. A more fundamental proof would follow the lines of the proof that sqrt(2) is irrational.
The answer requires a bit of mathematics, but goes like this:The product of any 2 rational numbers is a rational number.The product of any 2 irrational number is an irrational number.The product of a rational and an irrational number is an irrational number!Therefore simple logic tells us that there are more irrational numbers than rational numbers. There is a way to structure this mathematically, and I believe it is called an "Inductive Proof".Interesting !I'm going to say "No".I reason thusly:-- For every rational number 'N', you can multiply or divide it by 'e', add it to 'e',or subtract it from 'e', and the result is irrational.-- You can multiply or divide it by (pi), add it to (pi), or subtract it from (pi),and the result is irrational.-- You can take its square root, and more times than not, its square root is irrational.There may be others that didn't occur to me just now. But even if there aren't,here are a bunch of irrational numbers that you can make from every rational one.This leads me to believe that there are more irrational numbers than rational ones.-------------------------------------------------------------------------------------------------------There are infinitely many more irrationals than rationals; this was proved by G. Cantor (born 1845, died 1918). His proof is basically:The rational numbers can be listed by assigning to each of the counting numbers (1, 2, 3,...) one of the rational numbers in such a way that every rational number is assigned to at least one counting number;If it is assumed that every irrational number can be assigned to at least one counting numbers (like the rationals), then with such a list it is possible to find an irrational number that is not on the list; so is it not possible as there are more irrationals than there are counting numbers, which has shown to be the same size as the rational numbers, thus showing that there are more irrationals than rationals.
an irrational number is any real number that cannot be expressed as a ratio a/b, where a and bare integers, with b nonzero, and is therefore not a rational number.Informally, this means that an irrational number cannot be represented as a simple fraction. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals. As a consequence of Cantor's proof that the real numbers are uncountable (and the rationals countable) it follows that almost all real numbers are irrational.[1]When the ratio of lengths of two line segments is irrational, the line segments are also described as being incommensurable By Paul Philip S. Panis
No.
Yes. Google Cauchy's proof.
There cannot be a proof since your assertion is not necessarily true. sqrt(2)*sqrt(3) = sqrt(6). All three are irrational numbers.
No. The root word in "ir-RATIO-nal" is ratio. So an irrational number is one that cannot be written as a RATIO of two integers. Your number is a ratio of the two integers 3 and 8. Thus it is RATIO-nal. The examples of irrational number are infinity and certain square roots. Irrational number therefore cannot be expressed as a ratio a/b, where a and b are integers and b is non-zero. Informally, this means that an irrational number cannot be represented as a simple fraction. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals. As a consequence of Cantor's proof that the real numbers are uncountable (and the rationals countable) it follows that almost all real numbers are irrational. Perhaps the best-known irrational numbers are: the ratio of a circle's circumference to its diameter π, Euler's number e, the golden ratio φ, and the square root of two √2.
An ancient Greek mathematician in the era of Pythagoras is said to have been murdered by his fellow secret society members for discovering the (now well-known) proof that square root of 2 is irrational.
No integer is an irrational number. An irrational number is a number that cannot be represented as an integer or a fraction.All integers which are whole numbers are rational numbers.
It is a prime number that has only factors of itself and one therefore it is an irrational number like all prime numbers are.
There are uncountably infinite irrational numbers and their existence has been known for over 2500 years. In some cases, although their irrationality was suspected, rigorous mathematical proof took longer. Some notable events:sqrt(2): Pythagoreans (6th Century BCE).pi: Johann Heinrich Lambert (18th Century).e: Leonhard Euler (18th Century).In the late 19th Century, Georg Cantor proved that the number of irrationals is an order of infinity greater than the number of rationals.
Yes, any positive number is a number that doesn't have a (-) behind it (-20; -23.67; -45.45454...), and is not zero (0). Any repeating number (see 3rd negative example) is irrational, no matter what its sign. Irrational numbers also include numbers (decimals, specifically) that don't repeat, but don't stop. Numbers that don't terminate include pi. Pi, as it is, is proof of a positive irrational number.
The square root of a positive integer can ONLY be:* Either an integer, * Or an irrational number. (The proof of this is basically the same as the proof, in high school algebra books, that the square root of 2 is irrational.) Since in this case 32 is not the square of an integer, it therefore follows that its square root is an irrational number.