A formal proof for $2+2=4$ can be given (I used it as an exercise in a course once). The sought after solution depended on the definitions of: "$2$", "$4$" and "$+$" as they are commonly given in the context of Peano axioms.
It is clear that 2 + 2 = 4. It is also clear that applying the successor function on 1 yields the next number, i.e. 2, and this operation can be repeated infinitely. This method can be used to veri...
In elementary school, one learns that $2+2=4$ by experiment (putting two apples next to two other apples), and maybe also from some addition table to be memorized. But is there any approach that ...
The other interesting thing here is that 1,2,3, etc. appear in order in the list. And you have 2,3,4, etc. terms on the left, 1,2,3, etc. terms on the right. This should let you determine a formula like the one you want. Then prove it by induction.
Unary operators are all evaluated before binary operators, to keep things simple In the first case, $-2^2 = -4$. In the second case, $-2^2 = 4$. Most mathematical publications and mathematics forums will use the first case. However, some software (I believe Haskell like grammars are an example) use the second case.
Suppose the a discrete random variable X has a probability mass function p (x) = cx for x = 1,2,3,4,5 and p (x) = 0 for other values of x. (a) Find the value of the constant c such that p (x) is a valid
I am working on induction problems to prep for Real Analysis for the fall semester. I wanted proof verification and editing suggestions for part (a), and assistance understanding part (b). For part...
@MJD I have read the other post. My question is about the sequence and why the formula works and approaches the topic at a different angle as aposed to using the formula to extract the n. There are many similar looking but slightly different questions however I think it would be in best interest to keep this one. Thank you for your concern your effort is appreciated.
Hint: prove inductively that a product is $> 1$ if each factor is $>1$. Apply that to the product $$\frac {n!} {2^n}\: =\: \frac {4!} {2^4} \frac {5}2 \frac {6}2 \frac {7}2\: \cdots\:\frac {n}2$$ This is a prototypical example of a proof employing multiplicative telescopy. Notice how much simpler the proof becomes after transforming into a form where the induction is obvious, namely: $\:$ a ...
