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Is there a proof to show that if you multiply an irrational number times an irrational number you get a rational number?
There cannot be a proof since your assertion is not necessarily true.
sqrt(2)*sqrt(3) = sqrt(6). All three are Irrational Numbers.
What else can I help you with?
Why is 3 square root of 2 is irrational?
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.
Is an irrational number divided by a rational number always irrational?
Yes. The proof is easy. Let x be the irrational number and assume there exists some rational number r = a/b where a and b are integers (that's what it means to be rational). Now suppose x/r is a rational number. Then x/r = (b/a)x = c/d where c and d are some other integers. Since (b/a)x=c/d, then x = bd/ac which means that x itself is rational, but we assumed it was irrational. The contradiction proves that the assertion is wrong. An irrational divided by a rational must be irrational.
Are there more rational than irrational numbers?
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.
Is the 2 square root of 5 rational or irrational why?
It is irrational. The proof depends on the proof that sqrt(5) is irrational. However, judging by this question, I suggest that you are not yet ready for that proof. So, assume that sqrt(5) is irrational. Any multiple of an irrrational number by a non-zero rational isirrational. For suppose 2*sqrt(5) were rational that is 2*sqrt(5) = p/q for some integers p and q, where q is nonzero. then dividing both isdes by 2 gives sqrt(5) = p/(2q) where p and 2q are both integers and 2q in non-zero. But that implies that sqrt(5) is rational! That is a contradiction so 2*sqrt(5) cannot be rational.
How is the sum of a rational and irrational number irrational?
This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
Explain why the sum of a rational number and an irrational number is an irrational number?
Let R1 = rational number Let X = irrational number Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)
Can you subtract two rational numbers and get an irrational number?
Do you mean can we subtract one rational number from another rational number and get an irrational number as the difference? I'm not a mathematician, but I suspect strongly the answer is no. Wouldn't this imply that we can sometimes add a rational number to an irrational one, and get a rational number as a sum? That doesn't seem possible.Ans 2.It isn't possible. Proof :-Given two rational numbers, multiply the two denominators.Express each rational in terms of the common multiple.Algebraically add the numerators of the new rational numbers.Put this over the common multiple; there's the result expressed as a ratio.
If you add a rational and irrational number what is the sum?
an irrational number PROOF : Let x be any rational number and y be any irrational number. let us assume that their sum is rational which is ( z ) x + y = z if x is a rational number then ( -x ) will also be a rational number. Therefore, x + y + (-x) = a rational number this implies that y is also rational BUT HERE IS THE CONTRADICTION as we assumed y an irrational number. Hence, our assumption is wrong. This states that x + y is not rational. HENCE PROVEDit will always be irrational.
Is every integer an irrational number?
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.
Can it be demonstrated that there is a difference between the number of rational numbers and the number of irrational numbers?
Yes. Google Cauchy's proof.
Why is 3 square root of 2 is irrational?
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.
Is the square root of 32 an irrational or a rational 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.
Is an irrational number divided by a rational number always irrational?
Yes. The proof is easy. Let x be the irrational number and assume there exists some rational number r = a/b where a and b are integers (that's what it means to be rational). Now suppose x/r is a rational number. Then x/r = (b/a)x = c/d where c and d are some other integers. Since (b/a)x=c/d, then x = bd/ac which means that x itself is rational, but we assumed it was irrational. The contradiction proves that the assertion is wrong. An irrational divided by a rational must be irrational.
Can you add two irrational numbers to get a rational number?
Yes Yes, the sum of two irrational numbers can be rational. A simple example is adding sqrt{2} and -sqrt{2}, both of which are irrational and sum to give the rational number 0. In fact, any rational number can be written as the sum of two irrational numbers in an infinite number of ways. Another example would be the sum of the following irrational quantities [2 + sqrt(2)] and [2 - sqrt(2)]. Both quantities are positive and irrational and yield a rational sum. (Four in this case.) The statement that there are an infinite number of ways of writing any rational number as the sum of two irrational numbers is true. The reason is as follows: If two numbers sum to a rational number then either both numbers are rational or both numbers are irrational. (The proof of this by contradiction is trivial.) Thus, given a rational number, r, then for ANY irrational number, i, the irrational pair (i, r-i) sum to r. So, the statement can actually be strengthened to say that there are an infinite number of ways of writing a rational number as the sum of two irrational numbers.
Is the square root of 3 rational?
No, the square root of 3 is not rational.No. The square root of 3 is irrational.More generally: if p is a prime number then the square root of p is irrational and the proof of this fact mimics the famous proof of irrationality of the square root of 2.No - the square root of 3 is not rational, but the proof is too involved to post here.
Are there more rational than irrational numbers?
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.
Is the 2 square root of 5 rational or irrational why?
It is irrational. The proof depends on the proof that sqrt(5) is irrational. However, judging by this question, I suggest that you are not yet ready for that proof. So, assume that sqrt(5) is irrational. Any multiple of an irrrational number by a non-zero rational isirrational. For suppose 2*sqrt(5) were rational that is 2*sqrt(5) = p/q for some integers p and q, where q is nonzero. then dividing both isdes by 2 gives sqrt(5) = p/(2q) where p and 2q are both integers and 2q in non-zero. But that implies that sqrt(5) is rational! That is a contradiction so 2*sqrt(5) cannot be rational.
