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Let $C$ represent the value measured in degrees Celsius and $F$ denote the value measured in degrees Fahrenheit. Then we can relate $C$ to $F$ by this equation

$$C = \frac{5}{9}(F-32)$$

My question is what does it mean for a temperature increase of $1$ degree Fahrenheit? I thought that this number has the form $F+1$, but that's wrong, and such a temperature increase is represented simply by substituting $1$ into $F$. But if this is the case, then $C = -\frac{155}{9}$, yet it is supposed to be equivalent to a temperature increase of $5/9$ degree Celsius. So, what am I missing on this second point?

I summarize my two questions with this claim that a temperature increase of $1$ degree Fahrenheit is equivalent to a temperature increase of $\frac{5}{9}$ degree Celsius, and I want to determine whether it is true or false, but I don't understand what a temperature increase of $1$ degree Fahrenheit and a temperature increase of $\frac{5}{9}$ degree Celsius mean.

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    Yes, one degree F is smaller than one degree C – 2017-01-25
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    Did someone really tell you that a one degree Fahrenheit increase has anything to do with what happens when you substitute $1$ for $F$ in $\frac{5}{9}(F-32)$? Your first idea, while incomplete, is a better way to approach the problem. – 2017-01-25
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    @DavidK Yes. I initially began tackling the problem the way I described towards the beginning of my post, but the answer on the block discussed a particular value for $F$, viz. $1$, and so on, as I already described. I didn't bother writing out completely my method, because I wanted to see if I was on the right track by comparing my answer to someone else's. I knew mine was correct, but I thought I may have missed something simpler; but, apparently, I didn't. – 2017-01-26
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    An alternative method is to convert $1$ Fahrenheit to Celsius, then convert $0$ Fahrenheit, then subtract. But that's really just $F$ and $F+1$ where $F=0.$ It's a small variation on what you started doing. – 2017-01-26

4 Answers 4

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Let $F'=F+1$ denote the new temperature in Fahrenheit where $F$ is the old temperature. Let $C'$ and $C $ similarly represent the new and old temperature in Celsius.

Then we get $$C' =\frac {5}{9}((F+1)-32) =\frac {5}{9}(F-32) +\frac{5}{9} = C +\frac {5}{9} $$ Hope it helps.

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    This is correct, but I suspect the OP's difficulty may be more basic, because of the last sentence in the question. He may need to be told first that there are two situations, one with an old temperature and another with a new temperature, and that the word "increase" in the question refers to the difference between those two. – 2017-01-25
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Your statement that an increase of one degree Fahrenheit is equivalent to $\frac{5}{9}$ degree Celsius is accurate. The two temperature scales have a proportional relationship.

The equation you provide can be thought of as a function. You input a number that represents the temperature in Fahrenheit, and the result of the calculation is the temperature in Celsius.

An increase of one degree Fahrenheit means that the initial input is no longer $F$, but $F+1$. As Rohan shows, this is equal to $C+\frac{5}{9}$.

If you were to substitute 1 for F, that means the temperature is 1 degree F, not 1 degree warmer. I think that's what you were missing.

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Hint: Solve for $\Delta C$.

$$C = \frac{5}{9}(F-32)$$ $$C + \Delta C = \frac{5}{9}((F+1)-32)$$

Can you generalize this for a temperature increase of any magnitude?

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Considering the conversion to C as a function in F $\,\;C(F) = \frac{5}{9}(F-32)\,$, it is easy to see that it's a linear function, and its derivative is a constant $\frac{d\,C}{d\,F} = C' = \frac{5}{9}\,$.

Then, as for any linear function, $\Delta C = C' \,\Delta F\,$, so a change $\Delta F = 1$ of one degree F translates into a change of $\Delta C = \frac{5}{9} \cdot 1 = \frac{5}{9}$ degrees C.