Question

what is a rational number?

Answer 1

a number that can be put in the form:
\[\frac{a}{b}\]
where a and b are integers, and b isnt 0.

Answer 2

examples of rational numbers:
\[\frac{25}{5}, \frac{1}{3}, -\frac{18}{19}\]
Not rational numbers:
\[\sqrt{2}, \pi , e^1\]

Answer 3

so is 0 a rational number?

Answer 4

yes, 0 would be:
\[\frac{0}{1}\]

Answer 5

ok..so if b=0 then why is 0 an rational number? I dont undestand..

Answer 6

a is the one that equals 0, b = 1

Answer 7

the whole, "b cant be 0" thing is a safeguard to keep us from dividing by 0. You know, so we dont destroy the world.

Answer 8

OK..so if i have a \[\sqrt{7}\] why isnt this a rational number?

Answer 9

what level of math are you taking? there is a proof i could give, but if you are in say.....high school, it might be a little confusing.

Answer 10

I mean, not really confusing.....but you have to be used to mathematical proofs.

Answer 11

Im taking college intermediate algebra

Answer 12

hmm....i guess i can give it. If you understand, thats great, if not, dont worry bout it. Just know that the square root of a number that isnt a perfect square is always irrational.

Answer 13

ok

Answer 14

Alright, here we go. This is going to be a proof by contradiction. Im going to assume that the square root of 7 is a rational number, and whats going to happen is things arent going to make sense.
Assume:
\[\sqrt{7} = \frac{p}{q}\] Multiplying both sides by p and then squaring we get:
\[\sqrt{7} = \frac{p}{q} \Rightarrow q\sqrt{7} = p \Rightarrow 7q^2 = p^2\]
Now, p and q are integers. Using the Fundamental Theorem of Arithmetic, any integer can be prime factored, so im going to prime factor p and q.
\[p = a_1*a_2*\cdots *a_n\]\[q = b_1*b_2*\cdots *b_m\]
p has n primes, q has m primes.
Plugging that into our equation from earlier we have:
\[7q^2 = p^2 \Rightarrow 7(b_1*b_2*\cdots *b_m)^2 = (a_1*a_2*\cdots *a_n)^2\]
Now we count how many primes are on both sides of the equation. On the left side we have:
\[2m+1\]
The m comes from the m primes in q, the extra 1 is the 7.
On the right side we have:
\[2n\]
The n comes from the n primes in p.
The number (2m+1) is always odd for m = 1,2,3..... and the number (2n) is always even. There is no way an odd number can equal an even number. This is the contradiction. Therefore my assumption that square root of 7 is rational was false. It must be irrational.

Answer 15

ok..gotcha..I have \[\sqrt{16}\] which is 4..does this mean it isn't a rational number because if this was in factional form it would be 4 over 0? therefore since 0=b its not a rational number?

Answer 16

it would be rational because it would be \[\frac{4}{1}\]
anytime we have an integer by itself, we can just put a one beneath it to make it a rational number.

Answer 17

so any other square root that works out nicely will be a rational number.:
\[\sqrt{100} = 10 = \frac{10}{1}\]