# Maths proofWatch

#1
I am having trouble with this question, I am unsure as to where to go from here. any help would be appreciated.

https://gyazo.com/29ed7878bfd43d512cdc74f7e034e9a6
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4 weeks ago
#2
You need to substitute the odd numbers (1,3,5, etc) into the equation where the n is.

Hope I helped.
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4 weeks ago
#3
(Original post by MichaelCooper)
I am having trouble with this question, I am unsure as to where to go from here. any help would be appreciated.

https://gyazo.com/29ed7878bfd43d512cdc74f7e034e9a6
I would suggest explaining why you chose (2k + 1) in an exam.

Now look at the first two terms - take out a factor of 4 - what can you say about what's inside the bracket?
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4 weeks ago
#4
4n²+4n+1

First odd number: 1 so... 4x1²+4x2+1 = 9 which is one more than a multiple of 8.

Second odd number: 3 so 4x3²+4x3+1 = 49 which is one more than a multiple of 8.

And so on...

You're only proofing so a few times will do.
Last edited by Steve445; 4 weeks ago
0
#5
when I take out a factor of 4 from 4n2 +4n+1 I get:
4(n2+n)+1
I am confused as to how this helps as it shows its 1 more than a multiple of 4, not 8

(Original post by Muttley79)
I would suggest explaining why you chose (2k + 1) in an exam.

Now look at the first two terms - take out a factor of 4 - what can you say about what's inside the bracket?
0
4 weeks ago
#6
(Original post by Steve445)
4n²+4n+1

First odd number: 1 so... 4x1²+4x2+1 = 9 which is one more than a multiple of 8.

Second odd number: 3 so 4x3²+4x3+1 = 49 which is one more than a multiple of 8.

And so on...

You're only proofing so a few times will do.
Err no?
n is not odd its an integer, (2n+1) is odd.
Also substituting values does not give a proof, how many values would you have to sub?
Factor "4n" out of the first two terms on your second line, what do you get? Then think about how you get an 8 multiplied by something + 1.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#7
(Original post by MichaelCooper)
when I take out a factor of 4 from 4n2 +4n+1 I get:
4(n2+n)+1
I am confused as to how this helps as it shows its 1 more than a multiple of 4, not 8
Look at (n2+n)

If n is odd then n2 is ? so (n2+n) is ?

Repeat if n is even
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4 weeks ago
#8
(Original post by Steve445)
4n²+4n+1

First odd number: 1 so... 4x1²+4x2+1 = 9 which is one more than a multiple of 8.

Second odd number: 3 so 4x3²+4x3+1 = 49 which is one more than a multiple of 8.

And so on...

You're only proofing so a few times will do.
No it won't - a proof has to work for ALL values.
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4 weeks ago
#9
Oh no, ignore my post. I was thinking of something totally different.
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4 weeks ago
#10
Gonna have to look this topic on Mathswatch as for some reason I have forgotten it.
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4 weeks ago
#11
doesn't work for all odd values, how do you get 3?
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4 weeks ago
#12
(Original post by BrandonS03)
...
Edit your post - it is against the rules to post solutions - what the OP has done is fine anyway ....
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4 weeks ago
#13
4n+1 = 3
n is not an integer so the proof is not valid for 3^2
1
4 weeks ago
#14
(Original post by mqb2766)
4n+1 = 3
n is not an integer so the proof is not valid for 3^2
No your substituting values for n to show any value of n in this case of the statement OP provided is 1 more than a multiple of 8.
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4 weeks ago
#15
(Original post by BrandonS03)
No your substituting values for n to show any value of n in this case of the statement OP provided is 1 more than a multiple of 8.
4n+1 generates the odd numbers 1, 5, 9, 13, ... for integer n. It does not generate the odd numbers 3, 7, 11, ... which would not be covered by your proof.
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4 weeks ago
#16
(Original post by mqb2766)
4n+1 generates the odd numbers 1, 5, 9, 13, ... for integer n. It does not generate the odd numbers 3, 7, 11, ... which would not be covered by your proof.
I think your misunderstanding the question here. (4n+1)(4n+1) is an expression to show that the square of two odd numbers is 1 more than a multiple of 8. 4n+1 is an odd number for all values of n. So your trying to show that by substituting all values of n into the expression that it does equal 1 more than a multiple of 8. As demonstrated with n=3 in my workings. 169 is 1 more than 168 and 168 is a multiple of 8 (21x8).
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4 weeks ago
#17
(Original post by BrandonS03)
I think your misunderstanding the question here. (4n+1)(4n+1) is an expression to show that the square of two odd numbers is 1 more than a multiple of 8. 4n+1 is an odd number for all values of n. So your trying to show that by substituting all values of n into the expression that it does equal 1 more than a multiple of 8. As demonstrated with n=3 in my workings. 169 is 1 more than 168 and 168 is a multiple of 8 (21x8).
'fraid not. I want to show that 3^2 is 1 more than a multiple of 8. There is no integer value of n for which 4n+1 = 3. Hence your proof is not valid in this case, and the others mentioned above 7^2, 11^2, ...
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4 weeks ago
#18
(Original post by BrandonS03)
I think your misunderstanding the question here. (4n+1)(4n+1) is an expression to show that the square of two odd numbers is 1 more than a multiple of 8. 4n+1 is an odd number for all values of n. So your trying to show that by substituting all values of n into the expression that it does equal 1 more than a multiple of 8. As demonstrated with n=3 in my workings. 169 is 1 more than 168 and 168 is a multiple of 8 (21x8).
I'm a maths teacher - listen to what is being said - yours is not a general proof and I've already asked you to edit because you should not have posted it anyway.
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4 weeks ago
#19
You need an expression to show n is odd because right now your assuming n to be odd. 4n+1 guarantees that for all values of n its odd. So If n=7^2 then n=49 and substituting this into the expression of (4n+1)^2 we get 197^2 which is 38,809. 38,809-1 = 38,808. 38,808/8= 4851.
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4 weeks ago
#20
(Original post by BrandonS03)
You need an expression to show n is odd because right now your assuming n to be odd. 4n+1 guarantees that for all values of n its odd. So If n=7^2 then n=49 and substituting this into the expression of (4n+1)^2 we get 197^2 which is 38,809. 38,809-1 = 38,808. 38,808/8= 4851.
Last time ...
4n+1 is indeed odd ..., -3, 1, 5, 9, 13, ...
You've missed out half the odd numbers. There is no integer n for which you can analyse the square of 3 or 7 or 11 if you use the transformation 4n+1. It has to be something like 2n+1 or 2n-1 or ... to consider all odd numbers.

Edit - I don't know your level, but I'd encourage you to go over this as you're misunderstanding it and if you're doing gcse, you'd get marked down for it.
Last edited by mqb2766; 4 weeks ago
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