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Why is $(-1) \times (-1)=+1$ ? What is the intuitive concept ?
My second question :

How can I show that no triangular number can be of the form $3n-1$ ?

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  • $\begingroup$ If you are quit a debt you win money. If you do the opposite of going one step backwards then you go one step forward. Also, it is the only possibility in a ring once you admit the distributive law. $\endgroup$ Jul 24, 2013 at 18:04
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    $\begingroup$ Try to avoid posing unrelated questions as a single question. $\endgroup$
    – tomasz
    Jul 24, 2013 at 18:15

7 Answers 7

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First, let's establish a pattern that can be used to intuitively understand why $1 \times (-1) = (-1)$

$1 \times 3 = 3$

$1 \times 2 = 2$

$1 \times 1 = 1$

$1 \times 0 = 0$

Every time the factor goes down by one, the product goes down by 1 (which is the first factor here).

Extending the pattern:

$1 \times (-1) = -1$

$1 \times (-2) = -2$

and so on. So $a \times (-b) = - (a \times b)$

Well, we can apply the pattern to the other factor as well:

$(-1) \times 3 = -3$

$(-1) \times 2 = -2$

$(-1) \times 1 = -1$

$(-1) \times 0 = 0$

Every time the factor goes down by one, the product goes down by -1 (which is the first factor here), or in other words, goes up by 1.

Extending the pattern:

$(-1) \times (-1) = 1$

$(-1) \times (-2) = 2$

and so on. So $(-a) \times (-b) = a \times b$

Now for the second question:

A triangular number is of the form $n(n + 1)/2$. And of the form $3n - 1$ really means $-1 \pmod 3 \equiv 2 \pmod 3$.

If $n \equiv 0 \pmod 3$, $n(n + 1)/2 \equiv 0 \pmod 3$.

If $n \equiv 1 \pmod 3$, $n(n + 1)/2 \equiv 2/2 \equiv 1 \pmod 3$.

If $n \equiv 2 \pmod 3$, $n(n + 1)/2 \equiv 0 \pmod 3$.

None are equivalent to $2 \pmod 3$, so it can't be expressed as $3n - 1$.

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Multiplying Negatives has a nice answer involving a number line that may be useful.

So, by walking backwards, while facing in the negative direction, he moves in the positive direction.

As for triangular numbers, notice that you have to show they can't be congruent to 2 mod 3. Looking at the formula for these numbers from Wikipedia, the formula is $ \frac {n(n+1)}{2}$ which if you look at various values of n mod 3, you could compute the formula based on n which if you think of the products of consecutive numbers mod 3, you should see the answer I'd think.

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One way to think of multiplying by $-1$ is as a geometric operation of rotating by 180°. Note that this is exactly what happens on the real line, where a rotation by 180° just flips the sides.

In that case, what would it mean to rotate by 180° two times? It would be the same as rotating by 360°, which is the same as doing nothing. But this is the same as multiplying by 1.

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  • $\begingroup$ A picture would go a long way to make your answer even better. $\endgroup$
    – Yuriy S
    Jul 13, 2016 at 16:44
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Here are a couple of ways to look at it intuitively:


Positive and negative represent opposite directions:

When multiplying some value by a positive operand, we retain the direction of that value.

When multiplying some value by a negative operand, we invert the direction of that value.


Positive and negative represent the angles $0$ and $180$:

When multiplying two operands, we multiply their absolute values and add up their angles.

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For your first question, $(-1)\cdot n$ is the additive inverse of $n$, that is $(-1)\cdot n+n=0$. What is the additive inverse of $(-1)$?

For the second question, if we had the equality

$$\frac{m(m+1)}{2}=3n-1$$

This would mean that the quantity $m(m+1)$ is congruent to $4$ modulo $6$. Is that possible?

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One somewhat formal way to "see" it is by distributive law: $$0=-1\cdot 0=-1\cdot (1+-1)= -1+(-1)\cdot(-1)$$ so $-1+(-1)^2=0$ and $(-1)^2=1$. This may be circular, depending on how you define $-1$, but it might convince you...

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$$(-1) \times (2) = -2$$ $$(-1) \times (1) = -1$$ $$(-1) \times (0) = 0$$ $$(-1) \times (-1) = \cdots$$

Of course, maybe some of these aren't intuitive either!

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