Wikipedia:Reference desk/Archives/Mathematics/2013 December 28

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December 28

Number of integral coordinate (not a homework problem)

If A (3,4) and B is a variable point on the line |x|=6. If AB is less than or equal to 4, then the number of positions of B with integral co-ordinates is
(Ans: 5) 117.244.213.61 (talk) 05:26, 28 December 2013 (UTC)

I would want to solve this one by graphing it. I'd put a radius 4 circle around (3,4), and see where the circle and it's interior intersects the x = 6 and x = -6 lines. Then use the grid lines to find integer values of Y in that overlap. It should be visually obvious it won't hit the x = -6 line.

You could then confirm the distance between those points and (3,4) mathematically. If there was any question, you could also check points appearing to be just outside the circle. StuRat (talk) 15:15, 29 December 2013 (UTC)

But let's also take it on without graphing. In that case, the distance from (3,4) to the two nearest points on the lines is 3 to x = 6 and 9 to x = -6. Therefore, we can't use the x = -6 line.

Now let's apply the Pythagorean theory to the x = 6 line points:

D = sqrt(x2 + y2)

4 ≥ sqrt(9 + y2)

16 ≥ 9 + y2

7 ≥ y2

|2.64| ≥ y

y ≤ |2.64|

So, y is between -2.64 and +2.64. There are 5 integers in that range for y, which, when combined with x = 6 for all 5, gives us 5 points. StuRat (talk) 15:28, 29 December 2013 (UTC)

proofs on this

positive time positive equals positive (proof)

negative time negative equals positive (proof)

positive time negative equals negative (proof) 37.237.198.55 (talk) 09:31, 28 December 2013 (UTC)§

Let x, a, and b be any three real numbers defined by

x = (a)(b) + (-a)(b) + (-a)(-b). Here, 'x' is the sum of three different terms. Now, according to the rules of algebra, we can take 'b' as :common from first two terms or we can take '-a' as common from last two terms. Mathematically,

x = (a)(b) + (-a)(b) + (-a)(-b) = b(a + (-a)) + (-a)(-b) = b(0) + (-a)(-b) = (-a)(-b). Similarly,

x = (a)(b) + (-a)(b) + (-a)(-b) = (a)(b) + (-a)(b + (-b)) = (a)(b) + (-a)(0) = (a)(b). Therefore, we have x = (a)(b) and x = (-a)(-b). Now, :equating these two: (-a)(-b) = (a)(b). Hence, negative time negative equals positive.

The following pattern answers your questions more beautifully.

3 X -3 = -9

2 X -3 = -6

1 X -3 = -3

0 X -3 = 0

-1 X -3 = 3

-2 X -3 = 6

-3 X -3 = 9
Scientist456 (talk) 11:53, 28 December 2013 (UTC)

That proves nothing, it simply lists out cases using the rule. The real question is why do we have such a rule? There is no real proof for the simple reason that the rule was devised to make things work out in a reasonable sense. Saying that xy=0 if either x or y is less than 0 would be just as logical but wouldn't be as convenient or useful. For instance it is very convenient if xy plus xz is the same as x times y+z. If that is extended to negative numbers then we need x(y+(−y)) = 0 so x(−y) = −xy. Dmcq (talk) 11:59, 28 December 2013 (UTC)

I assume the question is about the set of real numbers. The real numbers constitute an ordered field by any modern construction of them. I don't know the details of proof that the real numbers (by any construction) actually is an ordered field (and frankly I'm not sure I want to know them, kind of boring mathematics - if there is such a thing), but I suspect that the three rules listed above follows directly from the definition (axioms) of an ordered field. YohanN7 (talk) 12:58, 28 December 2013 (UTC)

According to axioms

positive time positive equals positive

negative time negative equals positive

positive time negative equals negative

---

We could make axioms tend to negative direction

Exampe

Let (+a)(+b)=-ab , (-a)(+b)=+ab and (-a)(-b)=-ab

$(x-y)^{2}=-x^{2}-y^{2}+xy+xy=-x^{2}-y^{2}+2xy$

Let x=3 , y=4

$-9-16+2(3)(4)=-25+24=-1$

And

Let (+a)(+b)=+ab , (-a)(+b)=-ab and (-a)(-b)=+ab

$(x-y)^{2}=x^{2}+y^{2}-xy-xy=x^{2}+y^{2}-2xy$

$9+16-2(3)(4)=25-24=1$

What's true and proofs? — Preceding unsigned comment added by 37.237.199.165 (talk) 08:20, 29 December 2013 (UTC)

No, this will not work. That $(-a)(b)=-(ab)$ follows from the field axioms. Likewise with the others. (This doesn't even require the field to be ordered.) Sławomir Biały (talk) 14:45, 29 December 2013 (UTC)

We could make the three axioms you describe; it would simply be a notation reversal. However, they would contradict $1\cdot a=a$. Instead, -1 would be the multiplicative identity. Since we've already decided that 1 is the multiplicative identity, the current version of sign-multiplication is the only one that fits with associativity and distributivity (as noted by Sławomir Biały).--71.175.63.37 (talk) 15:49, 29 December 2013 (UTC)

Let x=y=1. We then have $0=(1-1)^{2}=1^{2}-1^{2}+2\cdot 1\cdot 1=2$ which is only true for fields (or rings) of characteristic 2. 95.115.187.53 (talk) 16:39, 29 December 2013 (UTC)

Sorry, I didn't notice you also want (+a)(+b)=-ab, but the effect is the same: let a=b=1, you get $1\cdot 1=-1$, thus 1=-1, add 1 to get 2=0 which is true exactly for rings and fields with characteristic 2. 95.115.187.53 (talk) 19:29, 29 December 2013 (UTC)

You do get $1\cdot 1=-1$ but this doesnt imply $1=-1$. Just that $-1$ is your multiplicative identity. — Preceding unsigned comment added by 84.92.32.38 (talk) 01:15, 30 December 2013 (UTC)