Commutative Law of Multiplication is a fancy way of saying when you multiply two numbers, it doesn’t matter which number you put down first and which number you put down second.

a * b = b * a

This basic law of arithmetic is taught in the second grade in elementary school. Yet it is very useful when you evaluate the relative merits between Traditional 401k, Roth IRA, and the new Roth 401k.

Blogger Trent writes the popular blog *The Simple Dollar*, which is one of the most successful personal finance blogs. Unfortunately Trent made the mistake of not recognizing the Commutative Law of Multiplication. In his post *The New Roth 401(k) Versus The Traditional 401(k): Which Is The Better Route?* he said Roth 401k is better even if the tax rate in the future is lower than the tax rate at present. His reasoning was

“Basically, by paying $2,800 a year now in extra taxes, Joe saves himself $14,000 a year in retirement.”

Wrong. It matters not how much tax you pay at different times. What matters is how much money you have left after all the taxes are paid. Sadly when more than one commenters pointed out the problem with Trent’s math, he still insisted that his math was correct. You would think a blogger writing about finance and investment should “get it,” but I guess not.

In case someone out there is still confused, here’s how the math works. Let t_{0} be the marginal tax rate now, and t_{1} be the marginal tax rate at retirement time. Suppose through successful investing, you are able to grow each dollar to $n when you are ready to retire. For each dollar you invest in a Traditional 401k, you will have $n before tax, and **n * (1 – t _{1})** after tax. In a Roth IRA or Roth 401k, for each dollar before tax, you pay tax first and have (1 – t

_{0}) dollars left after tax. Growing the money to the same degree, you will have

**(1 – t**when you are ready to retire. If the tax rate now (t

_{0}) * n_{0}) is the same as the tax rate at retirement time (t

_{1}), we have

n * (1 – t

_{1}) = (1 – t_{0}) * n

There, is the Commutative Law of Multiplication.

If the tax rate at retirement time is lower, t_{1} < t_{0}, Traditional 401k will be better than Roth 401k because the value on the left hand side is larger than the value on the right hand side. The opposite is true if the tax rate at present is lower, t_{0} < t_{1}.

Of course nobody knows what the future tax rates will be or whether they will be higher or lower than today’s. In choosing between a Traditional 401k and a Roth 401k, you just have to take a guess or do a little of both. For me, my money is on the Traditional 401k. I think the Roth 401k is a device for the current government to maximize its current revenue at the cost of robbing revenues from the future government. When the future government needs money, it will find ways to raise revenue including taxing on Roth withdrawals either directly or indirectly. The laws on Roth IRA and Roth 401k only say withdrawals from them *today* are not taxed. They don’t say withdrawals won’t *ever* be taxed. Tax laws can be changed by the legislature in the future.

Related Post: The Case Against Roth 401(k)

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Anonymous says

Thank you for this post. The Simple Dollar is, for whatever reason, a very popular sight despite the fact that (1) his posts usually lack content and (2) when he does have something, he often gets it wrong. I’ve followed his logic with squinted eyes since the get-go.

That said, I have enjoyed your postings. Relevant and factual. And you don’t feel some mandate to have six posts a day just to have six posts. Keeps it ‘simple’.

-Mike

kyle says

I disagree with your application of the commutative law to IRAs.

You’re forgetting the effect of contribution limits. By using a Roth IRA you can pay the taxes from your other income, effectively increasing the amount you can contribute. If you put the money in a traditional IRA, on the other hand, you may end up paying taxes the taxes from your distributions if you don’t other income to pay them.

Harry Sit says

Kyle – I hear you. The contribution limit is a problem, but only if you hit the limit. If contributing to a Roth IRA means contributing more, you can also increase the contribution to the traditional 401k instead. According to this study by Vanguard, only 10% of people max out their 401k. If you are one of the 10%, then yes, Roth means you can contribute more. For the other 90%, instead of contributing to Roth, they can also increase their 401k contributions. Then this math law applies.

Fred says

The common misconception is that the Roth option is better because “your earnings grow tax-free.” Looking closely at this claim:

If the future value of your investment is FV, and the present value is PV, then FV=PV(1+i)^y, where i is the interest rate per period and y is the number of periods the investment accumulates interest. For a traditional 401k, taxes are applied after interest has compounded, i.e. FV=(PV(1+i)^y)(1-t0). With the Roth option, taxes are applied to the present value, before interest accumulates, i.e. FV=(PV(1-t1))(1+i)^y. Because of the commutative property (not “law”) of multiplication, and assuming t1=t0, the FV is the same for both options. So although it seems that your earnings would grow tax-free for the Roth option, in fact, it’s more like the government taxes your earnings before you earn them.

Tom says

Nice post, agree that the simple dollar is often, well, too simple in its analysis.

However, isn’t your math wrong? What I mean is, the point you make is correct, you either pay tax upfront or on the back end, but you shouldn’t use two variables for taxation.

Saying n * (1 – t1) = (1 – t0) * n

is like saying n*(1-x)= (1-y)*n, which is not an example of the commutative property, its an equation with three variables.

bill says

but i think the author forgot about one fact (it is almost fact, if it is not 100% right), ie, $ is getting depreciated. so paying a lump-sum tax bill 30 years from now is better than paying NOW. in this way, even if tax rate are the same 30 years from now, i would argue that traditional is better than roth (with time value of money in consideration).