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I had a student ask what the relationship of algebra and geometry is.

does every algebraic concept theoretically have an equivalent geometric concept, even if it's impossible to draw/picture/visualize such as infinite dimensions?

for example, $x^n$ is the equivalent to 'volume' of the n-dimensional figure with dimensions whose length is $x.$

user29418
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    If you want a "yes or no" answer, then it is "no". – Somos Sep 05 '18 at 02:54
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    I think you need to clarify what you mean by a "concept". The only example you've given is a single algebraic expression. Are you asking whether every algebraic formula can be interpreted as some measurement of a geometric object? – mweiss Sep 05 '18 at 03:19
  • yes, mweiss. That makes it more clear. – user29418 Sep 05 '18 at 03:43

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The answer is no.

For example can you graph the function defined by $$f(x) =1 \text { if x is rational }$$ and $$f(x) =-1 \text { if x is irrational }$$

Could you graphically show that $$x^4-1= (x^2-1)(x^2+1)$$ or the binomial coefficient is $$ \frac {n!}{r!(n-r)!}?$$

Mohammad Riazi-Kermani
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  • doesn't that function look like a box (when you apply Lesbegue integration)? isn't the binomial coefficient just Pascal's Triangle? – user29418 Sep 12 '18 at 15:23
  • The box or the Pascal triangle are the oversimplification of deeper concepts. The box is not a geometrical box and the triangle is just a specific arrangement of numbers in that form. Of course I am not downgrading the importance of geometry in any way and I have used graphs in topology, discrete mathematics, real analysis and other subjects, but a graph is not a proof and does not work all the times. – Mohammad Riazi-Kermani Sep 12 '18 at 15:49