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- Who first defined truth as adæquatio rei et intellectus?
António Manuel Martins claims (@44:41 of his lecture quot;Fonseca on Signs quot;) that the origin of what is now called the correspondence theory of truth, Veritas est adæquatio rei et intellectus
- Difference between PEMDAS and BODMAS. - Mathematics Stack Exchange
Division is the inverse operation of multiplication, and subtraction is the inverse of addition Because of that, multiplication and division are actually one step done together from left to right; the same goes for addition and subtraction Therefore, PEMDAS and BODMAS are the same thing To see why the difference in the order of the letters in PEMDAS and BODMAS doesn't matter, consider the
- factorial - Why does 0! = 1? - Mathematics Stack Exchange
The theorem that $\binom {n} {k} = \frac {n!} {k! (n-k)!}$ already assumes $0!$ is defined to be $1$ Otherwise this would be restricted to $0 <k < n$ A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately We treat binomial coefficients like $\binom {5} {6}$ separately already; the theorem assumes
- Proof of $7^n-2^n$ is divisible by 5 for each integer $n \ge 1$ by . . .
Prove the following statement by mathematical induction $7^n-2^n$ is divisible by 5 for each integer $n \ge 1$ My attempt: Let the given statement be p (n) (1) $7-2
- matrices - How to multiply a 3x3 matrix with a 1x3 matrix . . .
I have 2 matrices and have been trying to multiply them but to no avail Then I found this online site and trying feeding it the values but yet no success - R' T is what i would like to do but
- Why is $\infty\times 0$ indeterminate? - Mathematics Stack Exchange
"Infinity times zero" or "zero times infinity" is a "battle of two giants" Zero is so small that it makes everyone vanish, but infinite is so huge that it makes everyone infinite after multiplication In particular, infinity is the same thing as "1 over 0", so "zero times infinity" is the same thing as "zero over zero", which is an indeterminate form Your title says something else than
- Prove that $1^3 + 2^3 + . . . + n^3 = (1+ 2 + . . . + n)^2$
HINT: You want that last expression to turn out to be $\big (1+2+\ldots+k+ (k+1)\big)^2$, so you want $ (k+1)^3$ to be equal to the difference $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\; $$ That’s a difference of two squares, so you can factor it as $$ (k+1)\Big (2 (1+2+\ldots+k)+ (k+1)\Big)\; \tag {1}$$ To show that $ (1)$ is just a fancy way of writing $ (k+1)^3$, you need to
- Good book for self study of a First Course in Real Analysis
Does anyone have a recommendation for a book to use for the self study of real analysis? Several years ago when I completed about half a semester of Real Analysis I, the instructor used "Introducti
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