Difference between revisions of "Talk:1562: I in Team"
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<sarcasm>There is an (annagram of) Randal in "People who don't understand how a proverb works" </sarcasm> No, seriously this is just cueball being a smart-ass. --[[Special:Contributions/162.158.91.230|162.158.91.230]] 08:53, 10 August 2015 (UTC) | <sarcasm>There is an (annagram of) Randal in "People who don't understand how a proverb works" </sarcasm> No, seriously this is just cueball being a smart-ass. --[[Special:Contributions/162.158.91.230|162.158.91.230]] 08:53, 10 August 2015 (UTC) | ||
− | There's no I in team, but there is an I in pie; there's an I in meat pie and meat is an anagram of team, so... | + | There's no I in team, but there is an I in pie; there's an I in meat pie and meat is an anagram of team, so... {{unsigned ip|141.101.99.82}} |
− | :There's a 999999 in pi. | + | :There's a 999999 in pi. {{unsigned ip|198.41.239.32}} |
:: Doesn't pi contain every possible number sequence though? [[Special:Contributions/162.158.91.235|162.158.91.235]] 11:17, 10 August 2015 (UTC) | :: Doesn't pi contain every possible number sequence though? [[Special:Contributions/162.158.91.235|162.158.91.235]] 11:17, 10 August 2015 (UTC) | ||
::: No. There is no evidence that pi includes an offset of pi. | ::: No. There is no evidence that pi includes an offset of pi. | ||
+ | :::There is no I in team, but there is meat... blessed meat :::Simpson drool:: {{unsigned|Cwallenpoole}} | ||
:::: ''finite'' sequence. the kate bush conjecture is unproven. {{unsigned ip|141.101.98.34}} | :::: ''finite'' sequence. the kate bush conjecture is unproven. {{unsigned ip|141.101.98.34}} | ||
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The arbitrariness of this saying was demonstrated considerably more elegantly in Jeffrey Rowland's Wigu: "There is no I in 'team', but there is in 'family'." [[Special:Contributions/198.41.242.93|198.41.242.93]] 11:56, 10 August 2015 (UTC) | The arbitrariness of this saying was demonstrated considerably more elegantly in Jeffrey Rowland's Wigu: "There is no I in 'team', but there is in 'family'." [[Special:Contributions/198.41.242.93|198.41.242.93]] 11:56, 10 August 2015 (UTC) | ||
This joke is not self-referential, it's metalingual. See https://en.wikipedia.org/wiki/Jakobson%27s_functions_of_language [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 13:10, 10 August 2015 (UTC) | This joke is not self-referential, it's metalingual. See https://en.wikipedia.org/wiki/Jakobson%27s_functions_of_language [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 13:10, 10 August 2015 (UTC) |
Revision as of 13:58, 10 August 2015
There is no I in team, but there is an M and an E.162.158.56.215 08:26, 10 August 2015 (UTC)
Check it out! there's "l" in "vowels"! --141.101.89.222 08:51, 10 August 2015 (UTC)
<sarcasm>There is an (annagram of) Randal in "People who don't understand how a proverb works" </sarcasm> No, seriously this is just cueball being a smart-ass. --162.158.91.230 08:53, 10 August 2015 (UTC)
There's no I in team, but there is an I in pie; there's an I in meat pie and meat is an anagram of team, so... 141.101.99.82 (talk) (please sign your comments with ~~~~)
- There's a 999999 in pi. 198.41.239.32 (talk) (please sign your comments with ~~~~)
- Doesn't pi contain every possible number sequence though? 162.158.91.235 11:17, 10 August 2015 (UTC)
- No. There is no evidence that pi includes an offset of pi.
- There is no I in team, but there is meat... blessed meat :::Simpson drool:: -- Cwallenpoole (talk) (please sign your comments with ~~~~)
- finite sequence. the kate bush conjecture is unproven. 141.101.98.34 (talk) (please sign your comments with ~~~~)
The arbitrariness of this saying was demonstrated considerably more elegantly in Jeffrey Rowland's Wigu: "There is no I in 'team', but there is in 'family'." 198.41.242.93 11:56, 10 August 2015 (UTC)
This joke is not self-referential, it's metalingual. See https://en.wikipedia.org/wiki/Jakobson%27s_functions_of_language Xhfz (talk) 13:10, 10 August 2015 (UTC)