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1. Let a and b be integers. Prove that if a|b, then a”|6" for all positive integers n.

1 Let A And B Be Integers Prove That If Ab Then A6 For All Positive Integers N class=

Sagot :

Solution

- a and b are integers. a | b means that integer a can divide integer b with no remainders.

- Let the Quotient of the division be k, so we can say:

[tex]\begin{gathered} a|b=k \\ \\ \text{ Put in an easier way, we have:} \\ \frac{b}{a}=k \\ \\ where, \\ k\text{ is an integer since }a\text{ directly divides b} \end{gathered}[/tex]

- Now, we are asked to find

[tex]a^n|b^n[/tex]

- Again, we can rewrite this as:

[tex]\frac{b^n}{a^n}[/tex]

- We can rewrite this expression using the law of exponents that says:

[tex]\frac{x^m}{y^m}=(\frac{x}{y})^m[/tex]

- Applying this law, we have:

[tex]\frac{b^n}{a^n}=(\frac{b}{a})^n[/tex]

- But we already know that

[tex]\frac{b}{a}=k[/tex]

- Thus, we have that:

[tex]\begin{gathered} \frac{b^n}{a^n}=k^n \\ \\ That\text{ is,} \\ a^n|b^n=k^n \\ for\text{ all positive integers n} \\ \\ k^n\text{ is an integer as well because }k\text{ is an integer.} \end{gathered}[/tex]

- Therefore, we have successfully proved the assertion