# Is it possible to express addition as repeated multiplication?

It is possible (and common when explaining multiplication) to express multiplication of positive integers as repeated addition like this:

$$4\cdot5 = 5+5+5+5=20$$

And a general expression would be this:

$$a\cdot b=\sum_{n=1}^ab$$

Is the reverse possible? Is it possible to express addition of positive integers as repeated multiplication?

$$a+b=\prod_{n=?}^??$$

Edit: This question is not a duplicate of How to define addition through multiplication? because my question does not put as strong limitations on what repeated multiplication means. A trivial answer would be $1\cdot(a+b)$, but is not so interesting. I found @barakmanos answer interesting enough.

• – User Jan 26 '15 at 20:51
• What are we allowed to use inside the $\prod$? Is a minus sign ok? – barak manos Jan 26 '15 at 20:58
• Let's say any real factor. – Jostein Trondal Jan 26 '15 at 21:01
• I was referring to arithmetic operations. For example, can we use $\prod\limits_{?}^{?}a-b$? – barak manos Jan 26 '15 at 21:03
• Choose any $n\in\mathbb{N}$ you like, then $a+b=\prod\limits_{i=1}^{n}\sqrt[n]{\frac{a^2-b^2}{a-b}}$ – barak manos Jan 26 '15 at 21:09

Choose any $n\in\mathbb{N}$ you like: $$a+b=\prod_{i=1}^{n}\sqrt[n]{\frac{a^2-b^2}{a-b}}$$
• This is recursive. $a - b$ is an addition. – rubik Jan 26 '15 at 21:48
• @rubik: So is ${a}\cdot{b}$, which also relies on addition. In fact, it is pretty obvious even in the example that OP gave within the question itself, of ${a}\cdot{b}=\sum\limits_{n=1}^{a}b$. Hence $a^b$ is also based on addition, and of course, $\ln({e^a}\cdot{e^b})$ (given in the link suggested in the first comment to the question), which is calculated as an infinite series consisting of several such arithmetic operations. You can say that pretty much every arithmetic operation is eventually based on a finite or infinite number addition operations. Please revoke your down-vote. – barak manos Jan 26 '15 at 22:02