MHB Defining & Proving Modulo Multiplication in $\mathbb{Z}_k$

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The discussion focuses on defining multiplication in the modular arithmetic system $\mathbb{Z}_k$ using an inductive approach. The proposed definition states that for non-negative integers, multiplication is defined as $x \cdot_k 0 = 0$ and $x \cdot_k (y+1) = x \cdot_k y +_k x$. Participants explore the validity of proving the equivalence $(x \cdot y) \mod k = (x \mod k) \cdot_k (y \mod k)$ through induction, questioning whether to apply induction on $y$. The base case and inductive step are outlined, demonstrating that both sides of the equation are equal under the defined operations. The conversation emphasizes the need for a clear inductive definition before proceeding with proofs.
mathmari
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Hey! :o

Based on the definition of $+_k$, I want to give an inductive definition for the multiplication $\cdot_k$ in $\mathbb{Z}_k$, such that for all $x\in \mathbb{N}_0$ and $y\in \mathbb{N}_0$ it holds that $$x\cdot_k y=(x\cdot y)\mod k$$

What is an inductive definition? (Wondering)

Do we write $x\cdot y=y+y+\ldots +y$ ($n$-times) and we apply each time the addition $+_k$ ? (Wondering)
I want to prove also by induction that for a $x\in \mathbb{N}_0$ it holds for all $y\in \mathbb{N}_0$ that $$(x\cdot y)\mod k=(x\mod k)\cdot_k (y\mod k)$$

We apply the induction on $y$ or not? (Wondering)
 
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mathmari said:
I want to prove also by induction that for a $x\in \mathbb{N}_0$ it holds for all $y\in \mathbb{N}_0$ that $$(x\cdot y)\mod k=(x\mod k)\cdot_k (y\mod k)$$

We apply the induction on $y$ or not? (Wondering)

Is the induction as follows? Base case: $y=1$ : $$(x\cdot 1)\mod k=x\mod k$$ The other side is $$ (x\mod k)\cdot_k (1\mod k)=(x\mod k)\cdot_k 1=x\mod k$$ So, they are equal. Inductive Hypothesis: We suppose that it holds for $y=n$ : $$(x\cdot n)\mod k=(x\mod k)\cdot_k(n\mod k)$$ Inductive step: We will show that it holds for $y=n+1$ : $$(x\cdot (n+1))\mod k=(x\cdot n+x)\mod k=(x\cdot n)\mod k+x\mod k \ \ \overset{\text{ Inductive Hypothesis } }{ = } \ \ (x\mod k)\cdot_k (n\mod k)+x\mod k$$ The other side is $$(x\mod k)\cdot_k ((n+1)\mod k)=(x\mod k)\cdot_k (n\mod k+1)\\ =(x\mod k)\cdot_k (n\mod k)+x\mod k$$ So, they are equal.

Is this correct? Could I improve something? (Wondering)
 
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mathmari said:
What is an inductive definition?
Is there a reason why you can't look this up in your lecture notes or textbook? Or see Wikipedia.

mathmari said:
Do we write $x\cdot y=y+y+\ldots +y$ ($n$-times) and we apply each time the addition $+_k$ ?
No inductive definition uses ellipsis.

mathmari said:
Is the induction as follows?
Until there is a definition, it makes no sense to prove anything.
 
An inductive definition for the multiplication $\cdot_k$ in $\mathbb{Z}_k$ is the following:
\begin{align*}&x\cdot_k 0=0 \\ &x\cdot_k (y+1)=x\cdot_k y+_kx, \ y\geq 0\end{align*}
right? (Wondering)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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