# How can Young's modulus be dimensionless, but still have units? [closed]

Young's modulus is the ratio of stress, which has units of pressure, to strain, which is dimensionless; therefore, Young's modulus has units of pressure.

From my reasoning if something is dimensionless, it should be a unitless ratio. So how come Young's modulus has a unit?

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Because that article has punctuation problems, will delete this question. Once you read this. –  GuySoft Jun 5 '13 at 12:17

## closed as too localized by Waffle's Crazy Peanut, Emilio Pisanty, Brandon Enright, dmckee♦Jun 5 '13 at 14:54

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Young's Modulus isn't dimensionless!

It says STRAIN is dimensionless (which is true).

SO Y = Stress/Strain = [Pressure]/[Dimensionless] = [Pressure]!

Young's Modulus has the same dimensions as that of pressure, which is:

$[M] [L]^{-1} [T]^{-2}$

And units of pressure, which is Pascal.

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Thanks, got confused there for a minute –  GuySoft Jun 5 '13 at 12:20
Hi mikhailcazi. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. This markup is very useful for understanding math equations and parameters. Please have a look here for an intro or our FAQ for more info. For example, $\theta$ results $\theta$, $\omega$ inserts $\omega$, etc. It's quite interesting. You can revise your post if you can ;-) –  Waffle's Crazy Peanut Jun 5 '13 at 14:22