Really Simple Induction qn... (1 Viewer)

Jashua_Long

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Prove true n=1 (trivial).

Assume true n=k

8^k - 1 = 7M

Prove true n=k+1

i.e. RTP

8^(k+1) - 1 = 7C

LHS = 8*8^k - 1
= 8 [ 7M +1 ] - 1 [ from assump]
= 8*7M +7
= 7[ 8M +1]
= 7C
 

Leffife

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Base Step:
Test true for n = 1
i.e. 8¹ - 1 = 7 (7 is divisible by 7, thus true)

Now, assume the function is true for n = k
i.e. (8^k) - 1 = 7k (k is just some integer)

Prove:
8(8^k) - 1
= 8(7k + 1) - 1 ........... from assumed before (look up) I rearranged (8^k) - 1 = 7k into 7k + 1 = 8^k
= 8(7k) + 8 - 1
= 8(7k) + 7
= 7(8k + 1)
= 7L for some integer L

So we got 8^(k + 1) - 1 = 7L, where we can state that the statement is divisible by 7.
 

Leffife

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Base Step:
Test true for n = 1
i.e. 8¹ - 1 = 7 (7 is divisible by 7, thus true)

Now, assume the function is true for n = k
i.e. (8^k) - 1 = 7k (k is just some integer)

Prove:
8(8^k) - 1
= 8(7k + 1) - 1 ........... from assumed before (look up) I rearranged (8^k) - 1 = 7k into 7k + 1 = 8^k
= 8(7k) + 8 - 1
= 8(7k) + 7
= 7(8k + 1)
= 7L for some integer L

So we got 8^(k + 1) - 1 = 7L, where we can state that the statement is divisible by 7.
 

Jashua_Long

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Base Step:
Test true for n = 1
i.e. 8¹ - 1 = 7 (7 is divisible by 7, thus true)

Now, assume the function is true for n = k
i.e. (8^k) - 1 = 7k (k is just some integer)

Prove:
8(8^k) - 1
= 8(7k + 1) - 1 ........... from assumed before (look up) I rearranged (8^k) - 1 = 7k into 7k + 1 = 8^k
= 8(7k) + 8 - 1
= 8(7k) + 7
= 7(8k + 1)
= 7L for some integer L

So we got 8^(k + 1) - 1 = 7L, where we can state that the statement is divisible by 7.
Souldn't use k twice, it makes things very confusing. I always use M or P, letters that don't come up elsewhere.
 

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