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Pascal's triangle extends infinitely downwards, meaning in only has two sides. Why then is it called a triangle if a triangle, by definition must have three sides?

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    $\begingroup$ Most illustrations of it don't show the entire thing, in which case it looks like a triangle. That's my guess. $\endgroup$ – basket Oct 25 '16 at 5:29
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    $\begingroup$ You're right, one should call it the Pascal Cone. $\endgroup$ – Michael Hoppe Oct 25 '16 at 5:33
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    $\begingroup$ Pascal's Cone? Can we get a Koch {snow}flake with that? $\endgroup$ – TripeHound Oct 25 '16 at 8:09
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    $\begingroup$ It's actually a triangle, Pascal just didn't have the time to make it shorter. $\endgroup$ – JiK Oct 25 '16 at 10:52
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    $\begingroup$ It has three sides, each with infinite length. $\endgroup$ – nitro2k01 Oct 25 '16 at 17:32
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enter image description here

"Excuse me! Who says my thing does not look a triangle?"

                                                                         ~ Blaise Pascal

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    $\begingroup$ It looks more like a square to me: :) $$ \begin{array}{cccccc} &&&&1\\ &&&1&&1\\ &&1&&2&&1\\ &1&&3&&3&&1\\ &&4&&6&&4\\ &&&10&&10\\ &&&&20 \end{array} $$ $\endgroup$ – Federico Poloni Oct 25 '16 at 12:58
  • $\begingroup$ .. and it can be extended in any direction, so it is a plane. $\endgroup$ – amI Oct 25 '16 at 22:10
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    $\begingroup$ Did Pascal really say the quoted line ? $\endgroup$ – A---B Oct 25 '16 at 22:18
  • $\begingroup$ No. Definitely not. $\endgroup$ – mtheorylord Oct 25 '16 at 23:47
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    $\begingroup$ @A---B no, it was Oscar Wilde. $\endgroup$ – hobbs Oct 26 '16 at 5:06
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I find Pascal's original form greatly preferable. Slightly modernized:

$$ \begin{array}{ll|llllllllll} \label{my-label} & & j & → & & & & & & & & \\ & & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline i & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ ↓ & 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \\ & 2 & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & & \\ & 3 & 1 & 4 & 10 & 20 & 35 & 56 & 84 & & & \\ & 4 & 1 & 5 & 15 & 35 & 70 & 126 & & & & \\ & 5 & 1 & 6 & 21 & 56 & 126 & & & & & \\ & 6 & 1 & 7 & 28 & 84 & & & & & & \\ & 7 & 1 & 8 & 36 & & & & & & & \\ & 8 & 1 & 9 & & & & & & & & \\ & 9 & 1 & & & & & & & & & \end{array} $$

This way, it's defined for all $(i, j) \in \mathbb{N}^2$. The identities are nicer, too:

$$T_{i,j} = T_{j, i}$$ $$T_{i,j} = \frac{(i+j)!}{i!j!}$$ $$T_{i+1,j+1} = T_{i+1,j} + T_{i,j+1}$$

Row sum:

$$\sum_{i+j=n}T_{i,j} = 2^n$$

The "hockey stick":

$$T_{i,j+1} = \sum_{0 \leq k \leq i}{T_{k,j}}$$

And finally, for fans of the binomial coefficient:

$$T_{i,j} = \binom{i+j}{i} = \binom{i+j}{j}$$

Of course, the higher-dimensional forms are easy too:

$$T_{\hat{x}} = \frac{(\sum_{x \in \hat{x}}{x})!}{\prod_{x \in \hat{x}}{(x!)}}$$ $$T_{\hat{x}} = T_{\sigma(\hat{x})}$$ $$T_{\hat{x}} = T_{x_1 - 1, x_2, ..., x_n} + T_{x_1, x_2 - 1, ..., x_n} + ... + T_{x_1, x_2, ..., x_n-1}$$

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As users basket and Michael Hoppe point out, truncations of Pascal's triangle to finite depth look like triangles, but the whole thing should really be called a cone. Naming conventions are horrible in math to begin with - this is the least of it!! Consider that orthogonal matrices have orthonormal columns, or that practically half of theorems with names attached are attributed incorrectly!

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    $\begingroup$ Everything is confusing until you learn it. $\endgroup$ – Mariano Suárez-Álvarez Oct 25 '16 at 6:45
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    $\begingroup$ Nobody is using the "mathematical" characteristics of a triangle or a cone in this case. So it doesn't matter. If you want to be serious, is that angle on the top acute, right or obtuse? And is it really on the top in the first place? $\endgroup$ – user23013 Oct 25 '16 at 11:01
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    $\begingroup$ @MarianoSuárez-Álvarez Then why is the group called orthogonal? $\endgroup$ – Federico Poloni Oct 25 '16 at 12:55
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    $\begingroup$ Federico makes a good point. The name "orthogonal group" is misleading, since it suggests these are precisely the matrices preserving orthogonality, but that is not a characterization of them. A matrix A in $M_n(\mathbf R)$ preserves orthogonality iff there is a $c \in \mathbf R$ such that $A\mathbf v \cdot A\mathbf w = c(\mathbf v\cdot \mathbf w)$ for all $\mathbf v$ and $\mathbf w$ in $\mathbf R^n$. Orthogonal matrices are only the special case $c=1$. Serge Lang points out the misleading nature of the term "orthogonal group" in his graduate algebra book. $\endgroup$ – KCd Oct 25 '16 at 13:05
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    $\begingroup$ And I would not be surprised if the explanation Mariano suggests is backwards: perhaps the term "orthogonal matrix" came first, so the name "orthogonal group" came from the fact that it consists of orthogonal matrices. $\endgroup$ – KCd Oct 25 '16 at 13:09

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