Digit Math Application: Proving that multiplying with 10**n puts n zeros at the end

Please read Digit Math: Introduction before you continue.

The problem

This is a very fundamental concept that we were taught when we were in junior schools. Now, think of it, what it says. If you add \(x\) ten times then you will get \(x\omega0\). If you add \(x\) hundred times then you will get \(x\omega00\); and so on. How do we know that this will always be true? We know it since we have never seen one violation of it, but, anyway I will try to prove it know to rest the uneasy souls.

The proof

Before I continue with this proof, I must make sure we understand another empirical rule.

When any number is multiplied by zero then the result is always zero. This is not hard to see why. When you multiply \(x\) by \(2\) then it is like adding \(x\) twice. When you multiply \(x\) by \(1\) then it is like adding \(x\) once. When you multiply \(x\) by \(0\) then it is like adding \(x\) zero times, which in other words is that we never added \(x\) in the first place so \(x\) never existed, so we had nothing, and that nothing is zero.

Back to proof.

Let the number \(a\omega b\) be multiplied by \(1\omega d\), where \(d\) is all all zeroes, \(n\) times. To use Bimultiplication forumula we need to make sure that \(b\) too has \(n\) digits. We can always partition a number such that \(b\) will always have \(n\) digits. For example if \(d = 000\) and \(a\omega b = 2\) then we make \(a = 000\) and \(b = 002\).


& a \omega b \times 10^d\\
&= a \omega b \times 1 \omega d\\
&= a.1 \omega (a.d + b.1) \omega b.d\tag{By Bimultiplication}\\
&= a \omega (d + b) \omega d\tag{Since d is all zeroes}\\
&= a \omega b \omega d\tag{Since d is all zeroes}

So, in the above equations we see that multiplying a number by \(10^d\) will give us the same number, but followed with \(d\) zeroes. (Proved)

Check out other applications of Digit Math

Link to list of other applications of Digit Math.

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