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### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Lesson 8: Topic C: Lesson 16: Rational and Irrational Numbers- Approximating square roots
- Approximating square roots walk through
- Approximating square roots
- Comparing irrational numbers with radicals
- Comparing irrational numbers
- Approximating square roots to hundredths
- Comparing values with calculator
- Comparing irrational numbers with a calculator
- 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

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# 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.

## Want to join the conversation?

- 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?(14 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.

Another example:

sqrt 5^2

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

16+3=19(25 votes)

- 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)

- 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?(3 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.(7 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)
- 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)

- Not very mathematical, but my way of thinking about this was "Of course, the number would still have all those non-repeating digits at the end." XD(2 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) - -5, 1/9, rational number why or why not(1 vote)
- Are these 2 numbers? or one mixed number? You have a comma in this middle which should not be there for a mixed number.

A rational number is any number that can be written as a fraction where the numerator and denominator are both integers.

-5 = -5/1. So it is rational.

1/9 is already in fraction form and meets the definition. So, it is rational.

-5 1/9 can be converted to an improper fraction, then it would satisfy the definition for a rational number.

Hope this helps.(2 votes)

## Video transcript

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.