I think about the difference between the two using differences instead of absolutes. That looks like this:

It’s kind of hard to do this calc:

F = [ (9/5) * C ] + 32

Or this one:

C = (5/9) * (F - 32)

I refer to those as absolute equations. You have to take into account the pesky offset everytime you want to convert. What if we drop it? This makes:

F = (9/5) * C = 1.8 * C

C = (5/9) * F ~= 0.6 * F

I refer to those as relative or difference equations because if you subtract a temperature from the other, you get the same thing:

F1 = [ (9/5) * C1 ] + 32

F2 = [ (9/5) * C2 ] + 32

F2 - F1 = [ (9/5) * C2 ] + 32 - { [ (9/5) * C1 ] + 32 }

= [ (9/5) * C2 ] - [ (9/5) * C1 ] + 32 - 32

= [ (9/5) * C2 ] - [ (9/5) * C1 ]

= (9/5) [ C2 - C1 ]

F2 - F1 = (9/5) (C2 - C1)

∆F = (9/5) ∆C

So, why is this useful?

Say you have a temperature in Celsius and want to go to Fahrenheit. Simply multiply that number in your head by 1.8 (or think of this as multiplying by 180° as in trig) and finally add to 32. So, 1 °C is (1 * 1.8) + 32 °F or about 34 °F.

Going the other way is a little bit weirder. I make approximations when going the other way by thinking of 180° and how that can be divided. So, 180°, 90°, 45°, etc. corresponds to 1.8 °F (1 °C), 0.9 (0.5 °C), 0.45 °F (0.25 °C), etc. I also approximate by choosing the nearest multiple of 5 or 10 °C (9 or 18 °F). So, 44 °F is between 41 °F (5 °C) and 50 °F (10 °C), closer to 41. It’s off by 3, which is about 3.6, which is 2 in Celsius world. This means 44 °F is about 7 °C.

Hope you get the gist! Celsius really is better. I remember this in a pinch:

10 °C = 50 °F

20 °C = 68 °F

30 °C = 86 °F

40 °C = 104 °F

50 °C = 122 °F

Etc.

The freezing temps are a little hard since you cross zero into negatives, but the extrapolation can help