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Question Number 166113 by mathls last updated on 13/Feb/22
prove that 1!=1
provethat1!=1
Commented by mathls last updated on 13/Feb/22
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Commented by MJS_new last updated on 14/Feb/22
there′s no proof. we define ∀n∈N^★ : n!=Π_(j=1) ^n j ⇒ 1!=Π_(j=1) ^1 j=1 also there′s no proof for 0!=1 we define 0!=1 because it makes sense.
there′snoproof.wedefine∀n∈N★:n!=∏nj=1j⇒1!=∏1j=1j=1alsothere′snoprooffor0!=1wedefine0!=1becauseitmakessense.
Answered by alephzero last updated on 14/Feb/22
n! = n ∙ (n−1)! Let n = 2 ⇒ 2! = 2 ∙ 1! ⇒ 2 = 2 ∙ 1! ⇒ 1! = 1
n!=n⋅(n−1)!Letn=2⇒2!=2⋅1!⇒2=2⋅1!⇒1!=1
Answered by Mathspace last updated on 14/Feb/22
1!=Γ(2)=∫_0 ^∞ t^(2−1) e^(−t) dt =∫_0 ^∞ t e^(−t) dt=[−te^(−t) ]_0 ^∞ +∫_0 ^∞ e^(−t) dt =[−e^(−t) ]_0 ^∞ =1 0!=Γ(1)=∫_0 ^∞ t^(1−1) e^(−t) dt =∫_0 ^∞ e^(−t) dt=[−e^(−t) ]_0 ^∞ =1
1!=Γ(2)=∫0∞t2−1e−tdt=∫0∞te−tdt=[−te−t]0∞+∫0∞e−tdt=[−e−t]0∞=10!=Γ(1)=∫0∞t1−1e−tdt=∫0∞e−tdt=[−e−t]0∞=1

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