# How can I explain horizontal shifts to a 12-year-old by analogizing with $\text{your money} = \text{my money} + 1?$

My 12-year-old cousin thinks this explanation is the most comprehensible, but she still can't relate the analogy with wealth inequality

If I say mymoney = yourmoney + 1 are you richer than me? No, you're poorer by a dollar! That's easy to see because when the variables have these names, of course you instantly and intuitively put yourself into the yourmoney frame of reference, and easily see that although 1 is being added to yourmoney, the result of that formula expresses my money and not yours!

to

The $$(x, y)$$ system must remain our frame of reference if we are to visualize the motion, and so what we do is rework the equations to isolate x and y:

\begin{align} s = x + 1 & \iff x = s - 1 \\ t = y + 3 & \iff y = t - 3 \end{align}

Now you can see that it's a move left and down and this is consistent with the minus signs.

The "backwards" issue arises when you put yourself into the frame of reference of the transformation, but continue using $$(x, y)$$ to think about that frame. You have to think about the backwards mapping: how do you recover $$(x, y)$$ from that other frame of reference?

When you see $$x + 1$$, you must not imagine that $$x$$ is moving. (This may be "brain damage" from working in imperative computer programming languages, especially ones like C and BASIC where the = sign is used for assignment: x = x + 1 moves coordinate x to the right.) Rather, $$x$$ is used as the input to a calculation that produces some other value. And so $$x$$ is that other value, minus 1.

1. To wit, she knows that $$s = x + 1 \implies$$ s is richer than x. But how does this imply a rightward shift?

2. Similarly, how does ($$t = y + 3 \implies$$ t is richer than y) $$\implies$$ downward shift?

• For what it's worth, 12 seems rather young to be working on this topic, unless she is very strong in math, in which case I would not have thought she'd have any trouble with it. I don't think I encountered this until I was 13 or 14, and it certainly wasn't at school, but rather when reading on my own about special relativity transformations (while knowing almost no algebra, but the popularizations I read in library books had pictures . . .). – Dave L Renfro Nov 22 '18 at 18:47
• – Jasper Nov 22 '18 at 19:14
• Does she understand that the convention is right = more, down = less? Your question 1 and question 2 seem to be quite different. Presumably, in your first question you mean that s has a rightward shift in the x-coordinate system, while your second question means that y has a downward shift in the t-coordinate system. – Acccumulation Nov 22 '18 at 20:44
• I think the problem is your emphasis on intuition. You keep banging on it like a hammer to a nail. But the path to entlightenment is from familiarity first, understanding later. Drill, drill, drill. Rather than looking for an aha of understanding. And FWIW, your writing about the stuff turned me off and seemed too technical. Just something about it made my eyes glaze over. – guest Nov 23 '18 at 6:29
• I think about $y = x-3$ as being like a tax of $3$ was imposed. The function $y=x$ has the same output as its input, but to get out the same values as before with the transformed function $y = x-3$ requires you to include an extra $3$ in your $x$-input, since it is going to get taxed off. The result of needing an extra $3$ for each $x$-input is graphically visible as moving $3$ to the right on the $x$-axis. – Benjamin Dickman Nov 24 '18 at 19:44

To be honest, it also took me some time reading the answers you linked to see what you are talking about, so I can fully understand how it might not be intuitive.

As you want to discuss the shift of the origin in a coordinate system, it might be best to draw said system and do lots and lots of examples. The simplest one I can think of right now would be something like $$y = x + a$$ and then figure out where the graph meets the $$x$$ axis, e.g. by drawing it. As it turns out, this is the case for $$x = -a$$, so kind of the same thing happens here as in what you are trying to discuss.
Once that is clear and well understood, you can explain how this is the same as moving the origin along the $$y$$ axis and then go into the topic you want to discuss.

Overall, I have to agree with the comments that this is rather advanced for a 12 (or 10?) year old to fully understand, beyond just learning the formula without thinking.

• But I'm not sure that coordinate shifts are key thing for the girl to be working now now. Maybe more important to build a base of familiarity with basic algebra and graphing first (draw lots of standard line graphs, learn the different forms, etc.) Then when she wants to look at line shifts, she does it with a firmer basis of initial understanding of unshifted line. – guest Nov 23 '18 at 14:49
• Please edit my post to clarify it. Sorry, but I don't understand how the rest of your answer explains the intuition? – NNOX Apps Nov 24 '18 at 18:02

Looking at the analogy in your question, suppose someone was confused about whether mymoney = yourmoney + 1 made me richer, or you richer, compared to mymoney = yourmoney. How would you help that person understand?

I think this is pretty clear: you would tell them to try some values. Ask them: If mymoney = 5, what is yourmoney in the two cases?

Now looking at your question, suppose someone is confused about whether y = x+1 moves a point to the left, or to the right, compared to y = x. How should you help that person understand?

So my answer to your question is that you should also tell them to try some values. Ask them: if y = 5, what is x in the two cases?

I don't know if I understood your question completely. You're asking for an intuitive explanation why $$\displaystyle \left\lbrace\begin{array}{l}s=x+5\\t=y+3\end{array}\right.$$

Consider an arbitrary point $$M(x,y)$$. $$x$$ and $$y$$ are the distance from the projections of $$M$$ on the axis to the origin. What are $$(s,t)$$? There's two ways to answer it.

### Active transformation

In this case $$(s,t)$$ are the coordinates of the point $$M'$$ image of $$M$$ by the translation of vector $$(5,3)$$. For example the image of $$A(2,2)$$ is $$A'(7,5)$$. Here we transform the point leaving the coordinate system intact.

### Passive transformation

I think what you're looking for is a passive transformation. In this case we don't find the image of $$M$$, we find a new coordinate system in which the coordinates of $$M$$ are $$(s,t)$$. This is equivalent to a translation of the coordinate system by a vector opposite to the previous vector, that is $$(-5,-3)$$.

Another way to see it is to translate $$M$$ into $$M'$$ using the vector $$(5,3)$$ leaving the coordinate system intact (active transformation until now) then translate back $$M'$$ (attached to the coordinate system) into $$M$$, that is translate $$M'$$ and the coordinate system by the vector $$(-5,3)$$.

Geometrically, $$x+5$$ is the horizontal distance between $$M$$ & $$(-5,0)$$ and $$y+3$$ is the vertical distance between $$M$$ and $$(0,-3)$$. So $$M$$ have coordinates $$(x+5,y+3)$$ iff the origin is $$(-5,-3)$$.