Hi,

I have to prove by MI that,x^5-1 is a multiple of 5.
My analysis is as follows:

1.If some no say x has to be a muliple of 5, then x/5 = n where n=1,2,3 etc.
2.In the prob above mentioned. while following the first step of MI,and substituing x with 1,we get the answer as 0.
Now,0 cannot be effectively called a "multiple" of 5.

Please send in suggestions tht wud help me in analysing ths problem.
Thanks.

0 IS a multiple of 5, but what are the conditions on x? If x is a multiple of 5 then obviously x^5-1 can't be a multiple of 5 too, and the Fundamental Theorem of Number Theory states that if something is false, then it is impossible to prove it, even by induction

Just wondering, what is the fundamental theorem of number theory that you mention?

Yeah! You need to put a condition on the values X can take.Else the statement cannot be proved!

>I have to prove by MI that, x^5 - 1 is a multiple of 5.

Sorry, but it fails at x = 2.
Is there a typo?

>My analysis is as follows:
>
>1.If some no say x has to be a muliple of 5, then x/5 = n
>where n=1,2,3 etc.

No, (x^5 - 1) is a multiple of 5, not x.

>2. In the prob above mentioned, while following the first
>step of MI,and substituting x with 1,we get the answer as 0.
>Now, 0 cannot be effectively called a "multiple" of 5.

But 0 is a multiple of 5.

Consider: "Is 18 a multiple of 6?"
Yes, because 18/6 = 3. That is, 18 is divisible by 6 with remainder 0.
Try that with "Is 0 a multiple of 5?"

Alternate approach: "Is 18 a multiple of 6?"
Yes, because 18 = 6k, for some integer k (namely, 3).
Is 0 = 5k for some integer k?