Discrete Math - Modular Arithmetic

hammonjj
Messages
32
Reaction score
0

Homework Statement


For which values of n≥2 does the implication:

axb=0 ⇔ a=0 or b=0

For some Zn (n should be a subscript)

NOTE: For the a x b, the x should be the x that has a circle around it. I didn't see that symbol in the "quick symbols" :)

Homework Equations


I know that this is true when n is a prime (answer in back of book), but I don't understand why. For example, if a=5 and b = 4, then that would also be 0 for n=5, right? Do I just not understand how to do modular arithmetic?

Thanks!
 
Physics news on Phys.org
What is your definition of that symbol? I would have interpreted as a "tensor product" but that doesn't really make sense here. I suspect it is a "special operation" your text is defining for this problem.
 
hi hammonjj! :smile:
hammonjj said:
For which values of n≥2 does the implication:

axb=0 ⇔ a=0 or b=0

I know that this is true when n is a prime (answer in back of book), but I don't understand why. For example, if a=5 and b = 4, then that would also be 0 for n=5, right?

ah, but 5 = 0 :wink:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top