Python, 3 bytes

min

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Takes a number string, outputs its smallest digit as a smallest character.

  The decimal starts

0.0123456789011111111101222222220123333333012344444401234555550123456666012345678801234567

Claim: This number c is transcendental

Proof:

Note that c is very sparse: almost all of its digits are zero. That's because for the nth digit for large n, there's high probability n has least one zero. Moreover, there are long runs of consecutive zeroes. We use existing results to show this means c is transcendental.

Following this math.SE question, let Z(k) represent the position of the k'th nonzero digit of c, and let c_k be that nonzero digit, a whole number between 1 and 9. Then, we express the decimal expansion of c, but only taking the nonzero digits, as as the sum over k=0,1,2,... of c_k/10^Z(k).

We use the result of point 4 of this answer: that c is transcendental if there are infinitely many runs of zeroes that are at least a constant fraction of the number of digits so far. Specifically, there must be an ε>0 so that Z(k+1)/Z(k) > 1+ε for infinitely many k. We'll use ε=1/9

For any number of digits d, take k so that Z(k) = 99...99 is made of d nines. There is such a k because this this digit is a 9, and so nonzero. Counting up from 99...99, these number all contain a zero digit, so this is the start of a large run of zeroes in c. The next nonzero digit isn't until Z(k+1) = 1111...11 with d+1 ones. The ratio Z(k+1)/Z(k) slightly exceeds 1+1/9.

This satisfies the condition for every number of digits d, proving the result.

Python, 3 bytes

min

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Takes a number string, outputs its smallest digit as a smallest character. For example, 254 gives 2. The decimal with these digits starts

0.0123456789011111111101222222220123333333012344444401234555550123456666012345678801234567

This is OEIS A054054.

Claim: This number c is transcendental

Proof: Note that c is very sparse: almost all of its digits are zero. That's because large n, there's high probability n has a zero digit, giving a digit min of zero. Moreover, c has long runs of consecutive zeroes. We use an existing result that states this means c is transcendental.

Following this math.SE question, let Z(k) represent the position of the k'th nonzero digit of c, and let c_k be that nonzero digit, a whole number between 1 and 9. Then, we express the decimal expansion of c, but only taking the nonzero digits, as as the sum over k=1,2,3,... of c_k/10^Z(k).

We use the result of point 4 of this answer by George Lowther: that c is transcendental if there are infinitely many runs of zeroes that are at least a constant fraction of the number of digits so far. Formally, there must be an ε>0 so that Z(k+1)/Z(k) > 1+ε for infinitely many k. We'll use ε=1/9

For any number of digits d, take k with Z(k) = 99...99 with d nines. Such a k exists because this digit in c is a 9, and so nonzero. Counting up from 99...99, these numbers all contain a zero digit, so it marks the start of a long run of zeroes in c. The next nonzero digit isn't until Z(k+1) = 1111...11 with d+1 ones. The ratio Z(k+1)/Z(k) slightly exceeds 1+1/9.

This satisfies the condition for every d, implying the result.

Python 2, 3 bytes

min

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Takes a number string, outputs its smallest digit as a smallest character.

The decimal starts

0.0123456789011111111101222222220123333333012344444401234555550123456666012345678801234567

Claim: This number c is transcendental

Proof:

Note that c is very sparse: almost all of its digits are zero. That's because for large n, there's high probability n has least one zero. Moreover, these zeroes come in long runs. We use existing results to show this means c is transcendental.

Following this math.SE question, let Z(k) represent the position of the k'th nonzero digit of c, and let c_k be that nonzero digit, a number between 1 and 9. Then, we express the decimal expansion of c, but only taking the nonzero digits, as as the sum over k=0,1,2,... of c_k/10^Z(k).

We use the result of point 4 of this answer: that c is transcendental if there are runs of zeroes that are at least a constant fraction of the number of digits so far. Specifically, there must be an ε>0 so that Z(k+1)/Z(k) > 1+ε for infinitely many k. We'll use ε=1/9

For any number of digits d, take k so that Z(k) = 99...99 is made of d nines. There is such a k because this place value has nonzero digit 9. Counting up from 99...99, these number all contain a zero digit, so this is the start of a large run of zeroes in c. The next nonzero place value isn't until Z(k+1) = 1111...11 with d+1 ones. The ratio Z(k+1)/Z(k) slightly exceeds 1+1/9.

This satisfies the condition for every number of digits d, proving the result.

Python, 3 bytes

min

Try it online!

Takes a number string, outputs its smallest digit as a smallest character.

The decimal starts

0.0123456789011111111101222222220123333333012344444401234555550123456666012345678801234567

Claim: This number c is transcendental

Proof:

Note that c is very sparse: almost all of its digits are zero. That's because for the nth digit for large n, there's high probability n has least one zero. Moreover, there are long runs of consecutive zeroes. We use existing results to show this means c is transcendental.

Following this math.SE question, let Z(k) represent the position of the k'th nonzero digit of c, and let c_k be that nonzero digit, a whole number between 1 and 9. Then, we express the decimal expansion of c, but only taking the nonzero digits, as as the sum over k=0,1,2,... of c_k/10^Z(k).

We use the result of point 4 of this answer: that c is transcendental if there are infinitely many runs of zeroes that are at least a constant fraction of the number of digits so far. Specifically, there must be an ε>0 so that Z(k+1)/Z(k) > 1+ε for infinitely many k. We'll use ε=1/9

For any number of digits d, take k so that Z(k) = 99...99 is made of d nines. There is such a k because this this digit is a 9, and so nonzero. Counting up from 99...99, these number all contain a zero digit, so this is the start of a large run of zeroes in c. The next nonzero digit isn't until Z(k+1) = 1111...11 with d+1 ones. The ratio Z(k+1)/Z(k) slightly exceeds 1+1/9.

This satisfies the condition for every number of digits d, proving the result.

Python 2, 3 bytes

min

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Takes a string, outputs a digit of its smallest character.

The first digits are

012345678901111111110122222222012333333301234444440123455555012345666601234567880123456789

I'm writing up the proof it's transcendental now.

Python 2, 3 bytes

min

Try it online!

Takes a number string, outputs its smallest digit as a smallest character.

The decimal starts

0.0123456789011111111101222222220123333333012344444401234555550123456666012345678801234567

Claim: This number c is transcendental

Proof:

Note that c is very sparse: almost all of its digits are zero. That's because for large n, there's high probability n has least one zero. Moreover, these zeroes come in long runs. We use existing results to show this means c is transcendental.

Following this math.SE question, let Z(k) represent the position of the k'th nonzero digit of c, and let c_k be that nonzero digit, a number between 1 and 9. Then, we express the decimal expansion of c, but only taking the nonzero digits, as as the sum over k=0,1,2,... of c_k/10^Z(k).

We use the result of point 4 of this answer: that c is transcendental if there are runs of zeroes that are at least a constant fraction of the number of digits so far. Specifically, there must be an ε>0 so that Z(k+1)/Z(k) > 1+ε for infinitely many k. We'll use ε=1/9

For any number of digits d, take k so that Z(k) = 99...99 is made of d nines. There is such a k because this place value has nonzero digit 9. Counting up from 99...99, these number all contain a zero digit, so this is the start of a large run of zeroes in c. The next nonzero place value isn't until Z(k+1) = 1111...11 with d+1 ones. The ratio Z(k+1)/Z(k) slightly exceeds 1+1/9.

This satisfies the condition for every number of digits d, proving the result.