- Proof: sum & product of two rationals is rational
- Proof: product of rational & irrational is irrational
- Proof: sum of rational & irrational is irrational
- Sums and products of irrational numbers
- Worked example: rational vs. irrational expressions
- Worked example: rational vs. irrational expressions (unknowns)
- Rational vs. irrational expressions
Proof: sum of rational & irrational is irrational
The sum of any rational number and any irrational number will always be an irrational number. This allows us to quickly conclude that ½+√2 is irrational. Created by Sal Khan.
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- now that we've proved that the sum of rational and irrational numbers is irrational, what about the sum of irrational and irrational numbers? And what about multiplying irrational numbers?(13 votes)
- An irrational number (added, multiplied, divided or subtracted) to another irrational number can be either rational OR it can be irrational..The test ( I just took it) shows examples of all these , that is, an irrational that is divided, subtracted, added, and multiplied to another irrational COULD be rational or irrational.
For instance, pi/pi. Well, anything divided by itself is 1. Thus, in this instance, it is rational. Another example, the sqrt of 5/ sqrt 5 is 1. Thus, it is rational.
If an irrational is taken to any root , for example, sqrt 5^2, if we raise it to the second power, it can be rational. Thus, the the sq root of 5 (which is really raised to the 1/2 power) and the exponent of 2 cancel each other out when you multiply them together, thus, you get 5, a rational number.
Or if you have an irrational that is sqrt5 (to the third root) raised to the third power, the 1/3 and 3 will cancel each other when you multiply the 3 and 1/3 to each other and it will become 5, a rational number.
Or, for instance, sqrt 5 times sqrt 5, both are irrational, but when you multiply them together, you get sqrt 25, which is 5.
Another example: Pi-Pi=0, which is a rational number.
Another example is if you multiply the conjugate of one term to another:
(4+sqrt3) (4-sqrt3) The irrational numbers cancel
16-4sqrt3 +4sqrt3 +sqrt 9
the 4sqrt3 cancels with the negative 4sqrt3.
The sqrt of 9 becomes 3
- Shouldn't we also prove that (mb-na) and nb have no common factor greater than 1?(6 votes)
- That step is not necessary for this proof since the values left over over after removing any common factor from (mb-na) and nb must still be integers and since they are integers, the proof still holds, that is, the outcome is not changed by removing common factors, even if they exist.
Good question by the way!(11 votes)
- Is it possible to have a number that is not rational but also not irrational?(4 votes)
- yes its 0, programming 1's and 0's, and infinity(3 votes)
- Is the sum of a rational number and irrational number always rational?(0 votes)
- This video has just proved that the sum of a rational number and an irrational number is irrational.(0 votes)
- Does anyone know why is it that irrational numbers cannot be expressed as a ratio of two integers? Take pi for example. Pi supposedly has an infinite number of digits. So it is impossible for us to know what all of these digits are and without knowing the numerator we cannot find the denominator.
Am I right?(2 votes)
- A rational number is defined as a number that can be written as a ratio of integers. This use of the term "rational" stems from the word "ratio".
We can prove that all rational numbers have repeating decimal expansions, and all numbers with repeating decimal expansions are rational.
Pi has been proven irrational (the proof is rather dense and requires analysis and calculus, so I won't go into it). Therefore, it's decimal expansion cannot be repeating, since if it repeated, pi would be rational.
The fact that pi's decimal expansion does not repeat is a derived property of it, not a defining one.(5 votes)
- So then if that's the case for adding and multiplying rationals and irrationals, would subtracting a rational from an irrational or an irrational from a rational and dividing a rational by an irrational or an irrational by a rational also result in an irrational number.(1 vote)
- Yes to all of those things.
Subtraction is just addition of the negative A negative rational is still a rational, and a negative irrational is still irrational.
Similarly division is just multiplication by the reciprocal (multiplicative inverse). The reciprocal of a rational is still rational (p/q -> q/p), and the reciprocal of an irrational is still irrational.(4 votes)
- a=1-root2 then how to find value of (a - 1/a)^3(0 votes)
- I will just process what I might try: a - 1/a, get common denominator to get (a^2 -1)/a, so note top is difference of perfect squares (a-1)(a+1)/a, substitute in (1 - √2 -1 ))(1 - √2+ 1)/(1 - √2) or (-√2)(2 - √2)/(1 - √2), multiply by (1 + √2)/(1 + √2) to get rid of root in denominator, (-√2)(2 - √2)(1 + √2)/(1 - 2). Next, start multiplying top out (-2√2 + 2)(1 + √2) or (- 2√2 - 4 + 2 + 2√2)/-1, which all goes down to -2/-1 or 2, (2)^3 = 8. This is assuming no math mistakes.(3 votes)
- Sum of a rational and irrational is irrational, but what about the sum of two irrationals?
(By irrationals, I mean pure surds which cannot be simplified.)
And also nothing like (√2) + (1 - √2) = 1(2 votes)
- Can sum of two irrational number be rational ?If yes can you please give example?(1 vote)
- Trivially, 2-√2 is irrational, so √2 + (2-√2) = 2 is an example.
An example more in the spirit of the question would be log(2) + log(5) = log(10) = 1
(Assuming you accept log(2) and log(5) are both irrational.)(2 votes)
- this video only shows that a rational+irrational=irrational but how about lets say 3/4+the sqaure root 64 how would u do that?(1 vote)
- Since both 3/4 and the square root of 64 (8) are rational numbers, the result is a rational number.(2 votes)
So I'm curious as to what happens if I were to take a rational number and I were to add it to an irrational number. Is the resulting number going to be rational or irrational? Well, to think about this, let's just assume it's going to be rational and then see if this leads to any form of contradiction. So let's assume that this is going to give us a rational number. So let's say that this first rational number we can represent as the ratio of two integers, a and b. Let's call this irrational number, let's just call this x. And their sum gives us another rational number. Well, let's express that as the ratio of two other integers, m and n. So we're saying that a/b plus x is equal to m/n. Well, another way of thinking about it-- we could subtract a/b from both sides and we would get our irrational number x is equal to m/n minus a/b, which is the same thing as n times b in the denominator. And then let's see. m/n is the same thing as mb over nb. So this would be mb. I'm just adding these two fractions. mb minus-- Let's see. a/b is the same thing as n times a over n times b. So minus n times a. All I did is I added these two fractions. I found a common denominator. So to make it clear, I multiplied this one b and b, and then I multiplied this one n and n then I just added these two things, and I got this expression right over here. So this denominator is clearly an integer. I have the product of two integers. That's going to be an integer. That's going to be an integer. And then this numerator, mb, is an integer. na is an integer. The difference of two integers. This whole thing is going to be an integer. So it looks like, assuming that the sum is rational, that all of a sudden we have this contradiction. We assumed that x is irrational, we're assuming x is irrational, but, all of a sudden, because we made that assumption, we're able to assume that we can represent it as this ratio of two integers. So this tells us that x must be rational. And that is the contradiction. That is a very large contradiction right over there. The assumption was that x is irrational. Now we got that x must be rational. So, therefore, this cannot be the case. A rational plus an irrational must-- so this is not right-- a rational plus an irrational must be irrational. Let me write that down. So a rational plus an irrational must be irrational.