Mixing
Proving injectivity for the set $A$ $C$ with $m-1$ elements.
If $A$ is a set with $m in mathbb{N}$ elements and $C subseteq A$ is a set with 1 element, then $A$ $C$ is a set with $m-1$ elements.
proof. 
I found this proof showing injectivity of $g$. When considering the case $i < k, j ge k rightarrow f(i) = f(j+1)$ we get a contradiction. Since we get a contradiction, does this mean that the implication is false? So $i < k, j ge k$ is true, while $f(i) = f(j+1)$ is false. If instead we said $f(i) ne f(j+1)$, would this make the implication true? Can we generalize this to all implications $(P rightarrow Q)$ that are false-so since $(P rightarrow Q)$ is false, it follows that ~ $(Prightarrow Q)$ is equivalent to ($P$ and ~$Q$) which is true. I feel like we can't do this, but I wanted to know concretely how we would proceed.
Also, having $i<k,j ge k$ already implies that $i ne j$ so would this $i<k,jge k$ case be redundant?
analysis proof-verification
asked Nov 20 '18 at 23:45
K.M
686312
add a comment |
If $A$ is a set with $m in mathbb{N}$ elements and $C subseteq A$ is a set with 1 element, then $A$ $C$ is a set with $m-1$ elements.
proof.
I found this proof showing injectivity of $g$. When considering the case $i < k, j ge k rightarrow f(i) = f(j+1)$ we get a contradiction. Since we get a contradiction, does this mean that the implication is false? So $i < k, j ge k$ is true, while $f(i) = f(j+1)$ is false. If instead we said $f(i) ne f(j+1)$, would this make the implication true? Can we generalize this to all implications $(P rightarrow Q)$ that are false-so since $(P rightarrow Q)$ is false, it follows that ~ $(Prightarrow Q)$ is equivalent to ($P$ and ~$Q$) which is true. I feel like we can't do this, but I wanted to know concretely how we would proceed.
Also, having $i<k,j ge k$ already implies that $i ne j$ so would this $i<k,jge k$ case be redundant?
analysis proof-verification
asked Nov 20 '18 at 23:45
K.M
686312
1
The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29
add a comment |
If $A$ is a set with $m in mathbb{N}$ elements and $C subseteq A$ is a set with 1 element, then $A$ $C$ is a set with $m-1$ elements.
proof.
I found this proof showing injectivity of $g$. When considering the case $i < k, j ge k rightarrow f(i) = f(j+1)$ we get a contradiction. Since we get a contradiction, does this mean that the implication is false? So $i < k, j ge k$ is true, while $f(i) = f(j+1)$ is false. If instead we said $f(i) ne f(j+1)$, would this make the implication true? Can we generalize this to all implications $(P rightarrow Q)$ that are false-so since $(P rightarrow Q)$ is false, it follows that ~ $(Prightarrow Q)$ is equivalent to ($P$ and ~$Q$) which is true. I feel like we can't do this, but I wanted to know concretely how we would proceed.
Also, having $i<k,j ge k$ already implies that $i ne j$ so would this $i<k,jge k$ case be redundant?
analysis proof-verification
asked Nov 20 '18 at 23:45
K.M
686312
If $A$ is a set with $m in mathbb{N}$ elements and $C subseteq A$ is a set with 1 element, then $A$ $C$ is a set with $m-1$ elements.
proof.
I found this proof showing injectivity of $g$. When considering the case $i < k, j ge k rightarrow f(i) = f(j+1)$ we get a contradiction. Since we get a contradiction, does this mean that the implication is false? So $i < k, j ge k$ is true, while $f(i) = f(j+1)$ is false. If instead we said $f(i) ne f(j+1)$, would this make the implication true? Can we generalize this to all implications $(P rightarrow Q)$ that are false-so since $(P rightarrow Q)$ is false, it follows that ~ $(Prightarrow Q)$ is equivalent to ($P$ and ~$Q$) which is true. I feel like we can't do this, but I wanted to know concretely how we would proceed.
Also, having $i<k,j ge k$ already implies that $i ne j$ so would this $i<k,jge k$ case be redundant?
analysis proof-verification
analysis proof-verification
asked Nov 20 '18 at 23:45
K.M
686312
asked Nov 20 '18 at 23:45
K.M
686312
asked Nov 20 '18 at 23:45
K.M
686312
asked Nov 20 '18 at 23:45
K.M
686312
asked Nov 20 '18 at 23:45
K.M
686312
686312
1
The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29
add a comment |
1
The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29
1
1
The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29
The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29
add a comment |
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The contradiction shows that $g(i)$ cannot be equal to $g(j)$ if $ilt kle j$. Therefore, all cases have been considered and $g$ is injective.
– John Douma
Nov 21 '18 at 1:29