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In this video: https://www.youtube.com/watch?v=zmhd8clDd_Y&t=1025s (called Five Stages of Accepting Constructive Mathematics)

at exactly 24:09 the lecturer says that it is possible to find a bijection between natural numbers and the real numbers in constructive mathematics. I wonder how one might do so.

Thanks.

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    You can find an answer to your question here : http://mathoverflow.net/questions/30643/are-real-numbers-countable-in-constructive-mathematics – 2017-01-05
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    He's saying there's an injection from the reals to the naturals, not a bijection (see my answer to an earlier question, http://math.stackexchange.com/a/2048058/59213 ). I think it's unlikely to be consistent that they are in bijection. Certainly the usual diagonal argument shows there is no surjection $\mathbb{N} \rightarrow \mathbb{N}^\mathbb{N}$, and assuming countable choice, no surjection $\mathbb{N} \rightarrow \mathbb{R}$. – 2017-01-05

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The lecturer, Andrej Bauer, has written at length about this issue in a MathOverflow post here.

Another key point is that, in minute 23 of the video, he emphasizes that he is describing things that can happen in some varieties of mathematics. As he says, "there are varieties where ... really bizarre things are permitted".

The term "constructive mathematics" refers to a collection of many different systems, which do not all prove the same facts. The lecture he is giving only lists things that can happen in some varieties of constructive mathematics.