charmedbeauty said:
x=1 (is the number assigned to the vairiable x, where 1 is a point on the real number line.)
x=1 (mod 5) (x can take on any real number, such that x-1= 5k for some integer k.)
Good. Your understanding is sound (except that x has to be an integer, not "any real number"). Basically, x = 1 means that x IS the actual number 1. x = 1 (mod 5) means (loosely) that when you divide x by 5, you get a remainder of 1. More formally, it means that x is of the form 5k + 1, where k is an integer (positive, negative or zero). This is the same as what you wrote (just rearrange).
x = 1 is clearly one number that satisfies the property x = 1 (mod 5), it occurs when k is zero. But there are an infinite number of integers that satisfy that property.
Now that we're done with the basics
, let's move on.
In this polynomial x^3 +x^2+x+2= 0 modulo 5
do I need to only find one case where this is true or all cases?
Let's call that polynomial P(x), shall we? P(x) = x^3 + x^2 + x + 2
You need to prove that in all cases where P(x) = 0 (mod 5), it also follows that x = 1 (mod 5).
The simplest way to do this is by exhaustion of cases, since there are so few.
First, enumerate the possibilities. x can be any integer, positive, negative or zero. But in every single case, x (mod 5) can only be 0,1,2,3 or 4. This is like stating that if you divide any integer by 5, you can only get the remainders 0,1,2,3 or 4.
Now we test each of those 5 cases in the original proposition. Remember that when we're doing the testing, x does NOT represent a single integer! It represents a whole class of integers (an infinite number) that give that certain remainder when dividing by 5 (loosely speaking). More properly, when we're testing (say) x = 2 (mod 5), we're testing all integers of the form 5k + 2, where k is an arbitrary integer. Each "case" of x that represents an infinite number of integers, is called a congruence class.
When we test, it may seem like we're acting like we can just substitute that number into the polynomial expression. For example, when we test the case x = 3 (mod 5), we're doing:
(3^3 + 3^2 + 3^1 + 2) (mod 5) = (27 + 9 + 3 + 2) (mod 5) = 41 (mod 5) = 1(mod 5).
Remember what I said, x is NOT (necessarily) the actual number 3 here (though one of the values that x can represent is 3). So how are we justified in doing this?
The answer is that in modulo math, we can multiply and take powers just like we do in normal math. So, if a = b (mod m), we can take nth powers of both sides and get a^n = b^n (mod m). We can also add like in normal math. For example, if a = b (mod m), then a + c = b + c (mod m).
Those properties (which actually need to be proved in an introductory class to modular arithmetic, but here I'm assuming them as axioms) allow us to just plug in those congruence classes almost like they were just single numbers. But we must be very clear that those are NOT single values of x we're plugging in, even though the calculation is exactly the same.
So test each of those values of x (mod 5). Which means (for example) writing down: "When x = 0 (mod 5), P(x) = 0^3 + 0^2 + 0 + 2 (mod 5) = 2(mod 5)", etc.
Once you've tested all 5 possible values of x(mod 5), you've exhausted all the cases.
You'll see that x = 1 (mod 5) is the ONLY case where P(x) = 0 (mod 5). Which means that P(x) = 0 (mod 5) implies x = 1 (mod 5), and you've proven the proposition you've set out to prove.
ie I can see that when x=1,then,
1+1+1+2=5 = 0 mod5
but surely it does not only apply when x=1 that x^3+x^2+x+2=0 mod5 is satisfied.
Im assuming I only need to find one case where x^3+x^2+x+2=0 mod5, since for any multiple of 5 this is true.
So why do I need to check other cases? If I have shown it is true for x=1 such that
1+1+1+2=5=0 mod5 and therefore 1-1=5(0) then is it not true for any x that satisfies x-1=5k that it will then satify x^3+x^2+x+2=0 mod5.??
I think I might be confused as to what is meant by x=1 mod5.
as my understanding this means that x-1= 5k for some integer k.
so when x=1 then 1-1=5(0) which is true.
But is this the right line of thinking?
I hope my rather long-winded post has made things much clearer for you.
Here's something else you may want to do to really SEE what's going on.
When x = r (mod 5), you know that x = 5k + r, right?
Try plugging in that value into P(x), i.e. work out P(5k+r). You don't even have to work modulo 5, just work it out in usual arithmetic fashion. Expand out everything using Binomial theorem.
Now, in that expression, identify which numbers are multiples of 5 and which are not. (Assume for the moment that r is not zero, i.e. that x is not a multiple of 5).
What happens when you take that expression modulo 5 (i.e. when you divide it by 5 and see what's the remainder)? Remember that all multiples of 5 "vanish" (become zero). What are you left with? Simplify that. Doesn't that just become the polynomial in terms of r, i.e. P(r)?
This is why if x = r (mod 5), you can just plug that r value into the polynomial like we did. In fact, hamsterman was essentially suggest you use this long method to prove the proposition, which might have been a good idea in hindsight, because it would've helped you to understand what's really going on.
