Induction Proof Verification
$begingroup$
I want to prove that $x^{2n}=(-x)^{2n}$. Here is my attempt, is this correct?
Step 1. $x^{2(1)}=(-x)^{2(1)}$ Which is true
Step 2. Assume: $x^{2k}=(-x)^{2k}$
Step 3. Proving $x^{2(k+1)}=(-x)^{2(k+1)}$
$x^{2(k+1)}= x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}x^{2}=(-x)^{2k+2}=(-x)^{2(k+1)} $
proof-verification induction
asked Jan 16 at 17:44
Jon dueJon due
948
$endgroup$
add a comment |
$begingroup$
I want to prove that $x^{2n}=(-x)^{2n}$. Here is my attempt, is this correct?
Step 1. $x^{2(1)}=(-x)^{2(1)}$ Which is true
Step 2. Assume: $x^{2k}=(-x)^{2k}$
Step 3. Proving $x^{2(k+1)}=(-x)^{2(k+1)}$
$x^{2(k+1)}= x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}x^{2}=(-x)^{2k+2}=(-x)^{2(k+1)} $
proof-verification induction
asked Jan 16 at 17:44
Jon dueJon due
948
$endgroup$
$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50
add a comment |
$begingroup$
I want to prove that $x^{2n}=(-x)^{2n}$. Here is my attempt, is this correct?
Step 1. $x^{2(1)}=(-x)^{2(1)}$ Which is true
Step 2. Assume: $x^{2k}=(-x)^{2k}$
Step 3. Proving $x^{2(k+1)}=(-x)^{2(k+1)}$
$x^{2(k+1)}= x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}x^{2}=(-x)^{2k+2}=(-x)^{2(k+1)} $
proof-verification induction
asked Jan 16 at 17:44
Jon dueJon due
948
$endgroup$
I want to prove that $x^{2n}=(-x)^{2n}$. Here is my attempt, is this correct?
Step 1. $x^{2(1)}=(-x)^{2(1)}$ Which is true
Step 2. Assume: $x^{2k}=(-x)^{2k}$
Step 3. Proving $x^{2(k+1)}=(-x)^{2(k+1)}$
$x^{2(k+1)}= x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}x^{2}=(-x)^{2k+2}=(-x)^{2(k+1)} $
proof-verification induction
proof-verification induction
asked Jan 16 at 17:44
Jon dueJon due
948
asked Jan 16 at 17:44
Jon dueJon due
948
asked Jan 16 at 17:44
Jon dueJon due
948
asked Jan 16 at 17:44
Jon dueJon due
948
asked Jan 16 at 17:44
Jon dueJon due
948
948
$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50
add a comment |
$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50
$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50
$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
correction
$x^{2(k+1)}=x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}(-x)^{2}=(-x)^{2k+2}=(-x)^{2(k+1)}$
same way you would show $(x+5)^{k+1}=(x+5)^{k}(x+5)$
this step doesn't prove anything $(-x)^{2k}x^{2}=(-x)^{2k+2}$
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
$endgroup$
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
correction
$x^{2(k+1)}=x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}(-x)^{2}=(-x)^{2k+2}=(-x)^{2(k+1)}$
same way you would show $(x+5)^{k+1}=(x+5)^{k}(x+5)$
this step doesn't prove anything $(-x)^{2k}x^{2}=(-x)^{2k+2}$
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
$endgroup$
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
add a comment |
$begingroup$
correction
$x^{2(k+1)}=x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}(-x)^{2}=(-x)^{2k+2}=(-x)^{2(k+1)}$
same way you would show $(x+5)^{k+1}=(x+5)^{k}(x+5)$
this step doesn't prove anything $(-x)^{2k}x^{2}=(-x)^{2k+2}$
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
$endgroup$
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
add a comment |
$begingroup$
correction
$x^{2(k+1)}=x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}(-x)^{2}=(-x)^{2k+2}=(-x)^{2(k+1)}$
same way you would show $(x+5)^{k+1}=(x+5)^{k}(x+5)$
this step doesn't prove anything $(-x)^{2k}x^{2}=(-x)^{2k+2}$
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
$endgroup$
correction
$x^{2(k+1)}=x^{2k+2}=x^{2k}x^{2}=(-x)^{2k}(-x)^{2}=(-x)^{2k+2}=(-x)^{2(k+1)}$
same way you would show $(x+5)^{k+1}=(x+5)^{k}(x+5)$
this step doesn't prove anything $(-x)^{2k}x^{2}=(-x)^{2k+2}$
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
answered Jan 16 at 18:16
Tariro ManyikaTariro Manyika
619
619
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
add a comment |
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
If you look at Step 2, I assume that $x^{2k} = (-x)^{2k}$, so I just plugged it in.
$endgroup$
– Jon due
Jan 16 at 20:17
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
thing is a lot of things are obvious but the point of a proof is to show it , you cant skip that step
$endgroup$
– Tariro Manyika
Jan 17 at 10:13
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
$begingroup$
Part of the inductive proof is to assume that it works for $k$, which is my step 2.
$endgroup$
– Jon due
Jan 17 at 11:54
add a comment |
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$begingroup$
Yes, it looks good for me.
$endgroup$
– Dog_69
Jan 16 at 17:50