hypertown@lemmy.world to Memes@lemmy.ml · edit-21 year agoA meme for math peoplelemmy.worldimagemessage-square72fedilinkarrow-up1956arrow-down137
arrow-up1919arrow-down1imageA meme for math peoplelemmy.worldhypertown@lemmy.world to Memes@lemmy.ml · edit-21 year agomessage-square72fedilink
minus-squareuberrice@feddit.delinkfedilinkarrow-up1arrow-down1·edit-21 year agoThis is confidently wrong. 3^0 is also 1. 2738394728^0 is also 1. Edit: just saw that technically you’re correct - sure. IF base 2, Exponent reduction equals to halving - dividing by 2. For x^y reducing y by one is equal to dividing by x, then we have the proof it always works.
minus-squareGlobulart@lemmy.worldlinkfedilinkarrow-up4·1 year agoBut that’s because for 3 the sequence is dividing by 3 not 2. 81, 27, 9, 3, 1, 1/3, 1/9, etc. 3^4, 3^3, 3^2, 3^1, 3^0, 3^(-1), 3^(-2), etc. You’re not always halving, but the method is the same and it sometimes helps people understand the concept more easily.
This is confidently wrong.
3^0 is also 1. 2738394728^0 is also 1.
Edit: just saw that technically you’re correct - sure.
IF base 2, Exponent reduction equals to halving - dividing by 2.
For x^y reducing y by one is equal to dividing by x, then we have the proof it always works.
But that’s because for 3 the sequence is dividing by 3 not 2.
81, 27, 9, 3, 1, 1/3, 1/9, etc.
3^4, 3^3, 3^2, 3^1, 3^0, 3^(-1), 3^(-2), etc.
You’re not always halving, but the method is the same and it sometimes helps people understand the concept more easily.