In David Graeber’s Debt: The First 5000 Years, he goes into a lengthy discussion about slavery, pointing out that a slave is “invisible”. They have no identity, and must perform their functions without notice, without drawing attention.

The same holds true for all sorts of other people who we’re taught “not to see”. From an early age, we’re taught not to stare at the disabled, not to go near poor people lest they try to grab us, or steal from us. In major cities, you simply do NOT make eye contact with anyone, ESPECIALLY the wretch begging on the street, or passing hat or cup on the subway. (In New York it is a crime to give money to someone panhandling on a subway or bus.) And so it goes for people in pain, who are drowning, who are dying. Don’t look. Don’t touch. Act like they’re not there. Invisible. Even when we do put a coin – or a bill- in the cup, we make sure not to make eye contact. It isn’t giving; it’s absolving ourselves, telling ourselves we’ve somehow helped, done our part. Yet, we never see the person, never see their poverty, their pain, them. Invisible.

I used to struggle with what, if anything, to give when I encountered a panhandler. Small change seemed like too little, perhaps even insulting. Enough for a meal might be good, but, how many meals for how many people, how often? Can’t help all of them, so now I must choose? Better to simply let all of them go hungry. That’s “fair”, right? And if I happen to have a $5 or $10 in my pocket, well, don’t I feel like the grand duke for giving someone that. And won’t that just make them the target of some other “thieving beggar”? Don’t want to feel haughty for giving alot, nor guilty for making them a target. And I sure as hell don’t want to look them in the eye and give them a glimpse of the discomfiture they’re causing me. Look straight ahead, keep walking, nothing to see here. Invisible.

I finally realized that while we no longer have slavery (as such) we deal with poverty – financial, spiritual, health (i.e. poor health) – in much the same way. We make the poor, the sick, the suffering, the dying. Invisible.

The solution is simple: SEE them. Put the coin, however small, in the cup. And when you do, don’t just drop it without stopping, never making eye contact. Stop, look at them, and say something to them, something that says, “I see you.” It doesn’t have to be in so many words, or in many words at all. “Good luck, mate!” is enough. I’ve never had a negative response to this. They always look up at me, smile, say, “God bless you”, or something along those lines. What you’ve given them isn’t money, or drugs, or booze, or food. You’ve recognized their humanity, their presence, their being. You make them … visible.

PI Day (3-14) 2019 Puzzle

This puzzle showed up in my twitter feed today, so I thought I’d post the solution as it makes use of both algebra and deductive reasoning. I’ll point out that the problem statement as seen in initial screen isn’t complete, and the solution he gives is to a somewhat different problem. What I solved is this:

Given the equation

\sqrt{PIE} = \sqrt{PI} + E

Where P, I, and E, are integers from 0 to 9, and P cannot be 0, find values for each of these that solve this equation.

The first thing I do is get all of the radicals on the same side …

\sqrt{PIE} - \sqrt{PI}  =  E

then square both sides and gather like terms, pulling out the common factor, PI …

PIE - 2PI\sqrt{E} + PI  =  E^2

PI(E - 2\sqrt{E} + 1)  =  E^2

Since we’re only considering integers, and E is the only term under the radical, our choices for it are limited to 0, 1, 4, and 9.

We can eliminate 0 right away. The only way the equation makes with E=0 is if P or I are also zero. P can’t be (by definition) and I can’t be since we’re not allowed to have both E and I set to zero.

Setting E = 1 makes the term in parentheses 0, which results in 0 = 1, which is of course absurd.

With E=9 we end up with 4 in the parenthetical, making the whole equation

4PI = 81

Since an product with an even factor must also be even, this rules out 9.


That leaves us with 4.  Plugging that in gives us PI = 16 which is entirely plausible. 

In fact, there are only two numbers left in our range that solve this: 2 and 8.  It doesn’t matter which you assign to P or I.  The math still works out.

Plugging these back in to the original equation, we see that

\sqrt{(2)(8)(4)} = \sqrt{(2)(8)} + 4

\sqrt{64} = \sqrt{16} + 4

8 = 4 + 4

The Original Puzzle

The puzzle I linked to at the top has one additional constraint: PIE is a three digit number, and PI is a two digit number. If you watch the video, the solution is almost pure deductive reasoning with little if any algebra at all. Oh, and the solution itself is completely different. In fact, one of the solution elements that doesn’t work here, does work in his version of the problem.