(New page: Category:MA453Spring2009Walther For this problem, does anyone know what the new conjecture is supposed to be? I thought it might just be that the order of r multiplied by the order o...)
 
 
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For this problem, does anyone know what the new conjecture is supposed to be?  I thought it might just be that the order of r multiplied by the order of s is NOT the order of rs, but I wasn't sure if there was another conjecture that could be made.
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For this problem, does anyone know what the new conjecture is supposed to be?  I thought it might just be that the order of r multiplied by the order of s is NOT the order of rs, but I wasn't sure if there was another conjecture that could be made. --[[User:Clwarner|Clwarner]] 21:14, 3 February 2009 (UTC)
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The conjecture I made was that <math>\scriptstyle gcd(r,s)=1\ \leftrightarrow\ \mid U(r)\mid*\mid U(s)\mid=\mid U(rs)\mid</math>.
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:--[[User:Narupley|Nick Rupley]] 05:01, 4 February 2009 (UTC)
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I made an equivalent conjecture:
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If r and s are relatively prime, then U(r) * U(s) = U(r*s).
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:--[[User:Ysuo|Yu Suo]]
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I discovered the same conjectur: U(r) * U(s) = U(r*s)
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:--[[User:Jzage|John Zage]]
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Obviously the conjecture they are leading you to is that U(r)*U(s)=U(r*s) but only if gcd(r*s)=1
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:--[[User:Ddesutte|Dane DeSutter]]

Latest revision as of 19:00, 5 February 2009


For this problem, does anyone know what the new conjecture is supposed to be? I thought it might just be that the order of r multiplied by the order of s is NOT the order of rs, but I wasn't sure if there was another conjecture that could be made. --Clwarner 21:14, 3 February 2009 (UTC)


The conjecture I made was that $ \scriptstyle gcd(r,s)=1\ \leftrightarrow\ \mid U(r)\mid*\mid U(s)\mid=\mid U(rs)\mid $.

--Nick Rupley 05:01, 4 February 2009 (UTC)

I made an equivalent conjecture: If r and s are relatively prime, then U(r) * U(s) = U(r*s).

--Yu Suo

I discovered the same conjectur: U(r) * U(s) = U(r*s)

--John Zage

Obviously the conjecture they are leading you to is that U(r)*U(s)=U(r*s) but only if gcd(r*s)=1

--Dane DeSutter

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