Can sbd help me with this- "A Few Words on Proofs"

E

Ellis Wan

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Hi this is a part of my math homework for the chinese new year hols- it's a reading scheme on "A Few Words about Proofs" could sbd help me plzplzplz?? tysm x(

Q1. Prove that X^(2)-1 is always divisible by 4 where X is an odd integer.

Solution 1: Since X is an odd integer, so the remainder when X is divided by 4 is either 1 or 3. If the remainder is 1, the remainder when X^2 is divided by 4 is 1x1=1. Hence X^(2)-1 is divisible by 4.

If the remainder is 3, the remainder when X^2 is divided by 4 is 3x3=9. Hence the remainder when X^(2)-1 is divided by 4 is the same as that of (9-1)/4 which is 0. Therefore, X^(2)-1 is always divisible by 4 where X is an odd integer.

Your comments on solution 1:

Your Solution:
​

Well actually, I have no idea what solution 1 is trying to say, but i can't really say that in the "comments". (really great start for the first q in the whole worksheet-.-) Should i just say how can (9-1)/4 even be 0??:confused: Anybody have a solution plzzz thanks
:*Ellis
 
Ellis Wan said:
Hi this is a part of my math homework for the chinese new year hols- it's a reading scheme on "A Few Words about Proofs" could sbd help me plzplzplz?? tysm x(

Q1. Prove that X^(2)-1 is always divisible by 4 where X is an odd integer.

Solution 1: Since X is an odd integer, so the remainder when X is divided by 4 is either 1 or 3. If the remainder is 1, the remainder when X^2 is divided by 4 is 1x1=1. Hence X^(2)-1 is divisible by 4.

If the remainder is 3, the remainder when X^2 is divided by 4 is 3x3=9. Hence the remainder when X^(2)-1 is divided by 4 is the same as that of (9-1)/4 which is 0. Therefore, X^(2)-1 is always divisible by 4 where X is an odd integer.

Your comments on solution 1:

Your Solution:
​

Well actually, I have no idea what solution 1 is trying to say, but i can't really say that in the "comments". (really great start for the first q in the whole worksheet-.-) Should i just say how can (9-1)/4 even be 0??:confused: Anybody have a solution plzzz thanks
:*Ellis
Click to expand...
The solution supplied is saying that if you divide an odd number 4 then the remainder is 1 or 3. To see this we can split up the odd numbers as 1, 5, 9, 13, 17... and 3, 7, 11, 15, 19,.... The first list has odd numbers that are all 1 more than the 3 times table. The 2nd list has odd numbers that are all 3 more than the 3 times table. So the remainders when you divide an odd number by is 1 or 3. Is this now clear?

The solution goes onto say that if when you divide an odd number by and the remainder is 1 then the remainder when x^2 is divided by 4 is 1. To see this not that x=4n+1. Then x^2 = (4n+1)^2= 16n^2 + 8n + 1 = 4(4n^2 + 2n) + 1. So the remainder is 1.


The solution then goes onto say that if when you divide an odd number by and the remainder is 1 then the remainder when x^2 is divided by 4 is 1. You prove this one

Now can you conclude that x^2 - 1 is divisible by 4.

For the record I do not like this proof at all, although it is valid. The way which Denis showed is the easiest way to go. (why do I support Denis???)
 
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