True or False: If a|b and a|(b + 1), then a = ±1.
[NOTE: Use this Theorem 3: If a|b and a|c, then a|(bx+cy) for any integers x and y.
PROOF: Since a|b, there is an integer u such that b = au.
Since a|c, there is an integer v such that c = av.
Substituting, bx + cy = aux + avy = a(ux + vy)
since ux+ vy is an integer, a|(bx+cy). QED.]
I get stuck with my prove. I'm trying to following theorem 3.
MY ANSWER: True
Proof: Since a|b, there is integer u such that b=au
by theorem 3 tell us that a| b+1 - b. so we have a = ±1.
please help.
if a|b and a|c, then a|(b-c) (why?)
so,
if a|(b+1) and a|b,
a|(b+1-b)
a|1
thank you so much steve
i get it.
a|b, b=ae
a|b+1, b+1=af
substituting, ae+1=af
1=a(e-f), we conclude that a|b+1-b
therefore a|1, a = ±1
To prove the statement "If a|b and a|(b + 1), then a = ±1", we can use the theorem you mentioned, Theorem 3. Let's walk through the proof step by step:
Given: a|b and a|(b + 1)
By the definition of divisibility, there exist integers u and v such that:
b = au (since a|b)
b + 1 = av (since a|(b + 1))
Now, we can use Theorem 3.
According to Theorem 3, if a|b and a|c, then a|(bx + cy) for any integers x and y.
Let's choose x = u and y = -v. Substitute these values into the equation bx + cy to get:
(b)(u) + (b+1)(-v) = au(u) + a(-v)(b+1) = a(u^2 - v(b+1))
Since u^2 - v(b+1) is an integer (as both u and v are integers), we have shown that a|(b(u) + (b+1)(-v)).
Now, we can conclude that a|(b + 1 - b), which simplifies to a|1.
Since a|1, there exist an integer k such that 1 = ak.
So, we have a = ±1, as 1 and -1 are the only integers that divide 1.
Thus, the statement "If a|b and a|(b + 1), then a = ±1" is true.