Question

Assume a and b are nonzero rational numbers and c and d are irrational numbers. For each of the following expressions, determine whether the result is irrational, rational, or both. Justify your answers.

Part A: ac + d

Part B: four b square root of three end square root plus c

Part C: b2(c + d)

Answer 1

@AihberKhan @mathmate @mathstudent55

Answer 2

@AlexandervonHumboldt2

Answer 3

A number can't be irrational and rational as the same time.
A rational number is a number the can be written as an integer/integer where the bottom integer is not 0.

So we can rewrite ac+d as
\[\frac{m}{n}c+d \\ \text{ where } m,n \in \mathbb{Z} -\{0 \} \text{ since } m,n \text{ are \not integers } \\ \text{ and } c \text{ and } d \text{ are irrational } \\ \text{ now what happens when you write the following as one fraction } \\ \frac{mc}{n}+d=? \\ \text{ do you wind up with an integer over an integer ? }\]

Answer 4

yes

Answer 5

I think ....

Answer 6

oh and I guess maybe those put the both in there because they want you to say well it can be an irrational but it could also be rational depending on the choices for a,b,c, and d....

anyways
what did you get when you combine the fractions above?

Answer 7

Im honestly kind of lost, im not sure what a,b,c, and d are

Answer 8

a, b are rational
c, d are irrational

rationals are numbers that can be written as integer/integer where bottom integer is 0
irrational is not rational (But still real)

Answer 9

Alrighty

Answer 10

It says in the beginning statesment that a and b are nonzero rational numbers.

Answer 11

\[\frac{mc}{n}+d=\frac{mc}{n}+\frac{dn}{n}=\frac{mc+dn}{n} \\ \text{ we defined } n \text{ as an integer} \\ \text{ but is } mc+dn \text{ an integer ? }\]
think about the possibilities for a sec...
what are the following equal to:
\[3 \sqrt{5}-3 \sqrt{5}=? \\ 5 \sqrt{5}-2 \sqrt{5}=?\]
I just chose some random possibilities for m,n,c and d

m and n are integers so they can be numbers in {...,-3,-2,-1,0,1,2,3,...}

sqrt(5) is just an irrrational number I just randomly picked to sub in for c and d

Answer 12

right an even though m and n are integers they can't be zero because we cannot have division by 0 (this is if n=0) and m/n can't be 0 ( so m can't be zero)

Answer 13

and I defined a as m/n

Answer 14

but anyways back to those two questions I just asked you

Answer 15

0

3\[3\sqrt{5}\]

Answer 16

that 3 in the middle doesn't belong there lol

Answer 17

right and 0 so what kind of number?
and 3 sqrt(5) is what kind of number?

Answer 18

answer being integer or not integer

Answer 19

Rational and irrational

Answer 20

im sorry
0 is an integer
and the other is not

Answer 21

ok 0 is an integer
and remember I defined n to be an integer
so 0/n is rational

but 3 sqrt(5) is not an integer
so 3 sqrt(5)/n is not rational

Answer 22

alrighty... are we working on part A? im kind of lost with the relation to the problem lol

Answer 23

we gave two examples that suggests the first could be rational or irrational depending on the choice for the variables in the expression given

Answer 24

we are trying to determine if the expression given can be written as an integer/integer or not

Answer 25

\[\frac{mc}{n}+d=\frac{mc}{n}+\frac{dn}{n}=\frac{mc+dn}{n} \\ \text{ we defined } n \text{ as an integer} \\ \text{ but is } mc+dn \text{ an integer ? }\]
the bottom is an integer because we defined n as an integer
we played with values for m,c,d, and n to see if the top was integer or not

Answer 26

Right, However the question asks if they are rational, irrational or both.
Do we have to configure if they are integers or not first?
Im kind of lost.

Answer 27

I understand for the most part what youre doing as long as im walked through yet im confused on the relation to this and the question asked :(

Answer 28

@freckles

Answer 29

I'm not sure what you are confused on

Answer 30

I defined a as m/n where m and n are integers
so I replaced a with m/n

Answer 31

\[\frac{mc+dn}{n}\]
we need to determine if we have integer over integer or not
we already know n is an integer
we must determine if mc+dn is an integer or not

if mc+dn is an integer then we have rational
if mc+dn is not an integer then we have not rational

Answer 32

I chose values for m,c,d, and n to show that it could be either rational or not rational depending on the selection

Answer 33

do you understand?

Answer 34

Not really.

Answer 35

a and b are non-zero rational numbers
c and d are irrational

The product of a rational number and an irrational number is always irrational.
The product of two irrational numbers may be rational or irrational.
The sum of two rational numbers is always rational.
The sum of two irrational numbers may be rational or irrational.

Part A.
ac + d
ac is a rational number multiplied by an irrational number. ac is irrational.
ac + d is an irrational number added to an irrational number.
The sum may be rational or irrational.

Part B.
\(4b\sqrt 3 + c\)
4 and b are rational. \(\sqrt 3\) is irrational, so \(4 b \sqrt 3 \) is irrational.
c is irrational.
The sum may be rational or irrational.

Part C.
\(b^2(c + d)\)
\(b^2\) is the same as \(b \times b\), so it is rational.
c + d may be rational or irrational.
\(b^2(c + d)\) may be rational or irrational

Answer 36

Here are examples of the statements I made above.

The product of a rational number and an irrational number is always irrational.
\(2\pi\), \(5\sqrt 6\), \(\dfrac{8}{9} \sqrt \pi \)
These are all examples of a rational number multiplied by an irrational number. They are all irrational.

The product of two irrational numbers may be rational or irrational.
\(\sqrt 2 \times \sqrt 2 = 2\): result is rational
\(\sqrt 3 \times \pi = \pi \sqrt 3\): result is irrational

The sum of two rational numbers is always rational.
0 + 1 = 1
5 + 10 = 15
1/2 + 1/6 = 7/6
All sums are rational.

The sum of two irrational numbers may be rational or irrational.
\(\pi + \sqrt 3\) is irrational.
\((4 + \pi) + (4 - \pi) = 4\) is rational.