[Someone]

No, it has to do with bothequally.

Changing a 0 bit to a 1 in an integer increases its value, and by the rule I gave, should also increase the value of the floating point number with the same bit pattern. It doesn’t matter whether that flipped bit is in the mantissa or the exponent.

That requires the use of the biased number in the exponent. In particular, the “all zeroes” bit pattern for the exponent must be the representation for the lowest possible exponent, and the “all ones” one that for the highest. It cannot be the ‘normal’ two’s complement representation of -1.

[saagarjha]

This is a different claim than the one above, which was about incrementing the least significant bit.

[Someone]

But it trivially follows from it. If operation foo (increasing x by one) makes something larger, repeated operation of foo (increasing x by 2^n) makes it larger, too.