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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
    Commented Oct 25, 2016 at 5:29
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    $\begingroup$ You're right, one should call it the Pascal Cone. $\endgroup$ Commented Oct 25, 2016 at 5:33
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    $\begingroup$ Pascal's Cone? Can we get a Koch {snow}flake with that? $\endgroup$
    – TripeHound
    Commented Oct 25, 2016 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
    Commented Oct 25, 2016 at 10:52
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    $\begingroup$ It has three sides, each with infinite length. $\endgroup$
    – nitro2k01
    Commented Oct 25, 2016 at 17:32

3 Answers 3

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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: :) $$ \begin{array}{cccccc} &&&&1\\ &&&1&&1\\ &&1&&2&&1\\ &1&&3&&3&&1\\ &&4&&6&&4\\ &&&10&&10\\ &&&&20 \end{array} $$ $\endgroup$ Commented Oct 25, 2016 at 12:58
  • $\begingroup$ .. and it can be extended in any direction, so it is a plane. $\endgroup$
    – amI
    Commented Oct 25, 2016 at 22:10
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    $\begingroup$ Did Pascal really say the quoted line ? $\endgroup$
    – user312097
    Commented Oct 25, 2016 at 22:18
  • $\begingroup$ No. Definitely not. $\endgroup$ Commented Oct 25, 2016 at 23:47
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    $\begingroup$ @A---B no, it was Oscar Wilde. $\endgroup$
    – hobbs
    Commented Oct 26, 2016 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. $\endgroup$ Commented Oct 25, 2016 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
    Commented Oct 25, 2016 at 11:01
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    $\begingroup$ @MarianoSuárez-Álvarez Then why is the group called orthogonal? $\endgroup$ Commented Oct 25, 2016 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
    Commented Oct 25, 2016 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
    Commented Oct 25, 2016 at 13:09

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