Proving Divisibility: Does a^n Divide b^n Imply a Divides b?

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In summary, the conversation discusses a mathematical exercise that aims to prove that if $a^n$ divides $b^n$, then $a$ must also divide $b$. The proof involves using the concept of greatest common divisor and ultimately concludes that $a$ does indeed divide $b$. There is some confusion about the notation used, but in the end, the solution is deemed correct.
  • #1
evinda
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Hey! (Wave)

I am looking at the following exercise:

Show that $a^n \mid b^n \Rightarrow a \mid b$.

According to my notes,it is like that:

Let $a^n \mid b^n$.

Let $d=(a,b)$

Then, $a=d \cdot a_1 \\ b=d \cdot b_1$

$$(a_1,b_1)=1$$

$$b^n=k \cdot a^n, \text{ for a } k \in \mathbb{Z}$$

$$d^n \cdot b_1^n=k \cdot d^n \cdot a_1^n \Rightarrow b_1^n=k \cdot a_1^n$$

Therefore, $$ a_1 \mid b_1^n=\underset{n}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \overset{(a_1,b_1)=1}{\Rightarrow} a_1 \mid \underset{n-1}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \Rightarrow a_1 \mid \underset{n-2}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \Rightarrow a_1 \mid b_1$$

So,we conclude that $b_1=l \cdot a_1, l \in \mathbb{Z}$

$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b \mid a \Rightarrow a \mid b$$

But...is it right?? (Thinking) We show that $(a_1,b_1)=1$ and then we conclude that $a_1 \mid b_1$... (Wasntme) :confused:
 
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  • #2
evinda said:
But...is it right?? (Thinking) We show that $(a_1,b_1)=1$ and then we conclude that $a_1 \mid b_1$... (Wasntme) :confused:

Hi! (Happy)

Then that must mean that $a_1=1$ doesn't it? (Thinking)
 
  • #3
I like Serena said:
Hi! (Happy)

Then that must mean that $a_1=1$ doesn't it? (Thinking)

Oh,yes! (Nod) So,the solution is right,or not?? (Thinking)
 
  • #4
evinda said:
$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b \mid a \Rightarrow a \mid b$$

evinda said:
Oh,yes! (Nod) So,the solution is right,or not?? (Thinking)

Almost.
But from $d \cdot b_1=l \cdot d \cdot a_1$ we cannot conclude that $b \mid a$. (Doh)
It's a good thing we can conclude that $a \mid b$! (Mmm)
 
  • #5
I like Serena said:
Almost.
But from $d \cdot b_1=l \cdot d \cdot a_1$ we cannot conclude that $b \mid a$. (Doh)
It's a good thing we can conclude that $a \mid b$! (Mmm)

Oh,sorry! (Blush)(Blush) I accidentally wrote it like that.I wanted to write:

$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b=l \cdot a \Rightarrow a \mid b$$

Thank you very much! (Smile)
 

FAQ: Proving Divisibility: Does a^n Divide b^n Imply a Divides b?

What is the meaning of the statement "a^n | b^n => a|b"?

The statement "a^n | b^n => a|b" means that if a number a is a factor of b raised to the nth power, then a is also a factor of b. In other words, if b can be divided by a certain number a n times, then b can also be divided by that number a at least once.

Can you provide an example to demonstrate this statement?

Yes, for example, let a = 2 and b = 8. We can see that 2^3 (a^n) is a factor of 8^3 (b^n) because 8 can be divided by 2 three times. Therefore, according to the statement, 2 (a) must also be a factor of 8 (b).

What are the implications of this statement in mathematics?

This statement has important implications in number theory, particularly in the study of divisibility and prime numbers. It helps us understand the relationship between factors and powers of numbers, and can also be applied in various mathematical proofs and calculations.

How is this statement related to the concept of prime numbers?

Prime numbers are numbers that are only divisible by 1 and themselves. This statement is related to prime numbers because it shows that if a number a is a factor of a prime number b raised to a certain power n, then a must also be a factor of b. In other words, if a number is a factor of a prime number's power, it is also a factor of the prime number itself.

Can this statement be extended to include negative exponents?

Yes, this statement can be extended to include negative exponents. In this case, the statement would be "a^n | b^n => a|b" for n > 0 and "a^n | b^n => a|b" for n < 0. The same logic applies, where if a number a is a factor of b raised to a certain power n, then a must also be a factor of b.

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