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# Argue that m x m is Not Equal to m + m

Here is what I have can you add anything to help this out?

You want to show that if m is a natural number (0, 1, 2, 3, ...), then m x m is not equal to m + m. You can do this simply by showing that it is not in at least one case (because if it's not true for at least one natural number, then it's not true for natural numbers in general.

Look at some cases:

m = 0
m x m = 0
m + m = 0
m x m equals m + m

m = 1
m x m = 1
m + m = 2
m x m is not equal to m + m

m = 2
m x m = 4
m + m = 4
m x m equals m + m

m = 3
m x m = 9
m + m = 6
m x m is not equal to m + m

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You could also start with the assumption that m x m = m + m, and try to find a contradiction:

m x m = m + m [assumption]
m x m = 2 x m [simplify right side]

if m is not equal to 0 ...

m = 2 [divide both sides by m (which we can do if m is not 0)]
We are left with the conclusion that the only number that m could be is 2. It can't be any other natural number

if m is equal to 0 ...
We know that m = 0. It can't be any other natural number.

Therefore, we didn't find a contradiction, but we did show that m must be 2 or 0 if m x m = m + m. The statement m x m = m + m is not true for all natural numbers.

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Another way to do this, using properties of natural numbers:

m x m = m + m [assumption]
m x m - m - m = 0 [subtract m + m from both sides]
m(m - 1 - 1) = 0 [property of distributivity]
m(m - 2) = 0 [simplify left side]
m = 0 OR m - 2 = 0 [properties of no zero divisors; if ab = 0, then a = 0 or b = 0 or both]

Again, either m = 0 or m = 2.

#### Solution Preview

The first section is sufficient to prove that m x m does not always = m + m. This is a proof by example, i.e. find a number that doesn't meet the equation. ...

#### Solution Summary

The solution assists with arguing that m x m is not equal to m + m.

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