[2024/05/13] Irrational powers

  • Show that it's possible a^b=c where a and b are irrational, and c is rational.

Sry for the gap I ran out of ideas.

11 points · 4 comments · view on lemmy.world

4 Comments

been_jamming@lemm.ee · 6 pts · 2y (1 reply)

e^(log 2) = 2 is rational

siriusmart@lemmy.world · 5 pts · 2y

that is simply genius

(i suppose it didnt come to me when i think of "irrational")

siriusmart@lemmy.world · 4 pts · 2y

this one got some table slams from my friends

Hint:

::: spoiler spoiler Find an example which satisfies the equation. :::


Solution:

::: spoiler spoiler https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd%20solutions/2024-05-13_irrational-powers.html :::

zkfcfbzr@lemmy.world · 1 pts · 2y

::: spoiler solution e^(i*π) = -1

Also, anything like a^(log(c) / log(a)), for positive rational c and irrational a, to generalize bean_jamming's answer

I also assert without proof that in the equation x^x = c, x is irrational for most rational values of c

I did start trying out stuff with sqrt(2), thinking back to the tower power problems, but didn't end up coming up with your solution while doing so ¯\_(ツ)_/¯ :::