Is temperature discrete

Question

Because an object's temperature is inversely proportional to the wavelength of blackbody radiation which it emits, physicists have theorized the existence of Planck temperature at around $1.4×10^{32}$ K.

Does this imply that temperature must be discrete? Meaning, the temperature could never be $1.0×10^{32}$ K, because that would imply a wavelength which is a non-integral multiple of the Planck length.

If it does imply this, does that mean that a system, if approaching the Planck temperature, and given more energy, would suddenly and radically change states from $0.7×10^{32}$ K to $1.4×10^{32}$ K? (perhaps much like a Bose Einstein condensate)

Pardon me if this question is bizarre, as I am not a physicist by profession (except maybe of the armchair variety).

Answer 1

Temperature cannot be discrete because it is an intensive variable:

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties are the temperature, refractive index, density....

In the simple formulation of the kinetic theory of gases,

$$T=\frac{m\overline{v^2}}{3 k_B},$$

where the temperature is directly connected with the mean of the square of the velocity of the molecules, and there is no discreteness in the formulation.

The Planck temperature is an estimate maximum attainable temperature with the physics we know, to be attained by matter given the constants we know, and has nothing to do with discreteness as normal temperatures are around $300\text{K}$.