Question

Prove that if a^2 is divisible by 7 then a is divisible by 7. heres my work thus far: http://imgur.com/UApos

Answer 1

I think you rationalize sqrt(7) so you get sqrt(7) on top and 7 bottom

Answer 2

\[\frac{a}{\sqrt{7}}=\frac{a \sqrt{7}}{7}\]

Answer 3

ah yea that would work

Answer 4

but 7 wouldn't be dividing only a

Answer 5

What do you know about integers and number theory?

If you know that every integer can a unique prime factorization, then there's an easy proof for your question. But I imagine you don't have this result. So what else do you know that we can use?

Answer 6

Im not sure whether I learned every integer can have a unique prime factorization,

Answer 7

Is the proof of that too large to contain within this one?

Answer 8

Actually, even without that strong result, I think you assume this:
Every positive p has A prime factorization,
\[ p = 2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5} ... \]
where the exponents a1, a2, a3 are positive or zero integers.

For example,
\[ 24 = 2^33^15^07^0 \ etc.\]

Answer 9

That being the case, there is a prime factorization of the integer a in this problem

\[ a = 2^{a_1}3^{a_2}5^{a_3}7^{a_4} ... \]
and therefore
\[ a^2 = 2^{2a_1}3^{2a_2}5^{2a_3}7^{2a_4} ... \]

Answer 10

But how do you know they all have the same power

Answer 11

Hmm, don't recall learning that

Answer 12

I'm not saying they have the same exponent. Just a exponent. Look at the example for 24.

Other examples,
\[ 200 = 2^35^2 \]
\[ 48 = 2^4 3^1 \]

Answer 13

Returning to the argument, given the form of a^2, then if \(a^2\) is divisible by 7, then the exponent of 7 in the factorization of a^2, i.e., \(2a_4\) has the property that \[ 2a_4 \geq 1 \]

Answer 14

Hence \( a_4 \geq 1/2 \). Now as \(a_4\) is an integer, \( a_4 \) is at least 1. I.e., a is divisible by 7.

Answer 15

Why does a4 must be an integer?

Answer 16

By construction. Every positive integer, p, has a factorization

\[ p = 2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}13^{a_6} ... \]
where the exponents \(a_i\) are integers, zero or positive.

Answer 17

Hmm

Answer 18

But for a^2, you can have 7^1
1 = 2a4
a4 = 1/2 ?

Answer 19

Again, in the example for 24 = 2^3 3^1, a_1 = 3, a_2 = 1, and all the other a_i are zero.

Answer 20

No. Read the argument carefully. If a^2 is divisible by 7 then 2a_4 is AT LEAST 1. It could be 17,003. But we know it's at least one.

And that means \( a_4 \geq 1/2 \). But as \(a_4\) is an integer, it must be that \(a_4 \geq 1\).

Answer 21

Why are you using a sub 4?

Answer 22

because a_4 is the exponent of 7.
\[ p = 2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}13^{a_6} ... \]

Answer 23

ok but you dont have to know anything about prime factorization for this though, it just helps

Answer 24

if a^2 is divisible by 7, we can conclude that a is greater than or equal to atleast 7

Answer 25

No knowledge of prime factorization is needed for that statement to be true correct?

Answer 26

I'm sure there are other proofs. I'm just giving you one.

Now, as for your statement: "if a^2 is divisible by 7, we can conclude that a is greater than or equal to at least 7". You need to show that. And then if it's true, how does it get you to your desired result?

Answer 27

oh i didnt mean 7

Answer 28

I meant 1

Answer 29

"if a^2 is divisible by 7 must be at least 1"

Answer 30

I dont have to show that do i?

Answer 31

No. By hypothesis of the question, a is a positive integer.

Answer 32

Ok so

Answer 33

from that we know a >=1/2

Answer 34

since by hypothesis of the question it must be a positive integer we know it must be at least 1

Answer 35

if it is 1 it is divisible by 7

Answer 36

Or is not right since a can be anything larger than 1

Answer 37

so it could be 2 which isnt divisible by 7