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Why does the sum of rational number and irrational numbers are always irrational?
Let your sum be a + b = c, where "a" is irrational, "b" is rational, and "c" may be either (that's what we want to find out). In this case, c - b = a. If we assume that c is rational, you would have: a rational number minus a rational number is an irrational number, which can't be true (both addition and subtraction are closed in the set of rational numbers). Therefore, we have a contradiction with the assumption that "c" (the sum in the original equation) is rational.
What makes a number rational?
A rational number is one that can be expressed as ratio(or fraction) of two integers: given integers a & b (with b ≠ 0), then the fraction a / b (or a ÷ b) will be a rational number. We need to specify that b not equal to zero, as division by zero is not defined (well not for most math applications).So irrational numbers such as pi, e, and square root of 2: cannot in any way be resolved into a fraction of two integers. All rational numbers can. Examples are:0 is rational, let a = 0, b = 1, or any other non-zero integer: 0/1 = 0Whole numbers are rational: 1/1 = 1; 2/1 = 2; 100/25 = 4, etc.Negative integers are also rational: (-4)/(2) = -2, or (20)/(-5) = -4All fractions (proper and improper) which have integers in the numerator, and non-zero integer in denominator, as well as mixed numbers are rational {positive and negative} will be rational numbers.
What operation is the set of positive rational numbers not closed?
subtraction. Let's take 1/2 and subtract 3/4 which is great than 1/2 so the answer is negative and hence not a positive rational.
Why is the sum of a rational numbers and an irrational number is irrational?
Let x be a rational number and y be an irrational number.Suppose their sum = z, is rational.That is x + y = zThen y = z - xThe set of rational number is closed under addition (and subtraction). Therefore, z - x is rational.Thus you have left hand side (irrational) = right hand side (rational) which is a contradiction.Therefore, by reducio ad absurdum, the supposition that z is rational is false, ie the sum of a rational and an irrational must be irrational.
Is square root 36 rational?
Hmm, let me see ... 6. Since I'm not still writing decimals, yes, it's rational.
Why is the product of any two rational numbers a rational number?
Rational numbers are represented in the form of p/q , where p is an integer and q is not equal to 0.Every natural number, whole number and integer can be represented as rational number.For example take the case of integer -3, it can be represented in the form of p/q as -3/1 and q is not equal to zero, which means that rational numbers consist of counting numbers, whole numbers and integers.Now, what will be the result of product of any two rational numbers?Let us take the case of two rational numbers which are x/y & w/z, their product is equal toxw/yz, which is a rational number because multiplication of x and w results in an integer and also multiplication of y and z results in an integer which satisfies the property of rational numbers, which is in the form of p/q.So, product of any two rational numbers is a rational number.
Can you multiply two rational numbers and come up with an irrational number?
No Explanation Suppose you could. Let x be an irrational number. Then take two rational numbers a,b and let ab=x Since a and b are rational, so is their product. So if x=ab, we have written a rational number as a ratio, namely ab/1 which is a contradiction. SO we conclude we cannot. A simpler answer might be just to say that rational numbers are closed under multiplication so it is impossible to get an irrational as the product of rationals.
Would the difference of a rational number and a rational number be rational?
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
Why does the sum of rational number and irrational numbers are always irrational?
Let your sum be a + b = c, where "a" is irrational, "b" is rational, and "c" may be either (that's what we want to find out). In this case, c - b = a. If we assume that c is rational, you would have: a rational number minus a rational number is an irrational number, which can't be true (both addition and subtraction are closed in the set of rational numbers). Therefore, we have a contradiction with the assumption that "c" (the sum in the original equation) is rational.
Why is the product of a non - zero rational number and an irrational number is irrational?
Let q be a non-zero rational and x be an irrational number.Suppose q*x = p where p is rational. Then x = p/q. Then, since the set of rational numbers is closed under division (by non-zero numbers), p/q is rational. But that means that x is rational, which contradicts x being irrational. Therefore the supposition that q*x is rational must be false ie the product of a non-zero rational and an irrational cannot be rational.
What are rational numbers that are also integers?
All integers {..., -2, -1, 0, 1, 2, ...} are rational numbers because they can be expressed as p/q where p and q are integers. Let p equal whatever the integer is and q equal 1. Then p/q = p/1 = p where p is any integer. Thus, all integers are rational numbers.
What makes a number rational?
A rational number is one that can be expressed as ratio(or fraction) of two integers: given integers a & b (with b ≠ 0), then the fraction a / b (or a ÷ b) will be a rational number. We need to specify that b not equal to zero, as division by zero is not defined (well not for most math applications).So irrational numbers such as pi, e, and square root of 2: cannot in any way be resolved into a fraction of two integers. All rational numbers can. Examples are:0 is rational, let a = 0, b = 1, or any other non-zero integer: 0/1 = 0Whole numbers are rational: 1/1 = 1; 2/1 = 2; 100/25 = 4, etc.Negative integers are also rational: (-4)/(2) = -2, or (20)/(-5) = -4All fractions (proper and improper) which have integers in the numerator, and non-zero integer in denominator, as well as mixed numbers are rational {positive and negative} will be rational numbers.
What operation is the set of positive rational numbers not closed?
subtraction. Let's take 1/2 and subtract 3/4 which is great than 1/2 so the answer is negative and hence not a positive rational.
How would you put rational in a sentence?
Rational means to be reasonable or to have reason. Let us be reasonable/rational. He was rational in his desition.
Is 2.51 rational?
Any number that you can completely write down using digits is rational.Let's try to write 2.51 completely with digits:2.51That's it! We're pretty sure that 2.51 is rational. The official math definition of a rational number is: A number that can be writtenas a fraction made of whole numbers. Let's try and write 2.51 as a fractionmade of whole numbers:251/100We did it again! 2.51 is definitely rational.
What are the rational numbers from 1 to 100?
Rational numbers are fractions. There are infinitely many fractions between 1 and 100. You cannot list them all.But numbers like 1/2 and 1/3 are rational and so are ones like 7 which is 7/1.If you give me any two rational numbers, say 6/8 and 7/8, I can find a rational number in the middle. Let's just right 6/8 as 12/16 and 7/8 as 14/16 then 13/16 is in the middle of those two. I can do that again with 13/16 and 14/6 by writing them as26/32 and 28/32 and 27/32 in the middle.I am sure you can see how I can keep doing this forever. This illustrates how between any two rational numbers there is always another. In fact, I just pick the number in the middle of the two, but there are many others between any two rational numbers. We say the rational numbers are a dense subset of the real numbers.
Is a rational number divided by an irrational number always irrational?
No. If we let x be irrational, then 0/x = 0 is a counterexample. However, if we consider nonzero rational numbers, then our conjecture is true. We shall prove this by contradiction. Suppose we have nonzero rational numbers x and y, and an irrational number z, such that x/z = y. Since z is not equal to 0, x = yz. Since y is not equal to 0, x/y = z. Since x/y is a quotient of rational numbers, x/y is rational. Therefore, z is rational, contradicting our assumption that z was irrational. QED.
