- #1

- 287

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Hint: If not, it must factor as (ax+b)(cx+d) with a,b,c,d in Q. Show that this is impossible.

So I got this far:

ac =1

ad+bc=0

bd=1

I'm not sure how to go further than this.

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- Thread starter chaotixmonjuish
- Start date

- #1

- 287

- 0

Hint: If not, it must factor as (ax+b)(cx+d) with a,b,c,d in Q. Show that this is impossible.

So I got this far:

ac =1

ad+bc=0

bd=1

I'm not sure how to go further than this.

- #2

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Hint: Square.

- #3

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I asked my prof if just saying it factors to (x-i)(x+i)...however he said it needs to be general. Hence why I have the ac, ac+db, and bd. I'm not sure how to 'generalize' the proof.

- #4

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- #5

- 287

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ad=-bc

a=-bc/d

sub into the first equation

-bc^2=1

c^2=-1/b

^ This is impossible in Q

- #6

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No, b could be negative. Try substituting in the other direction.

- #7

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well i'm saying that c^2 can't equal a negative number in Q, it would imply an i somewhere

- #8

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Yes, but if b is negative, then -1/b will be positive, which is not an issue.

- #9

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hmm...so I'm not sure what you mean then.

- #10

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By substitute in the other direction, did you mean:

a=1/c and b=1/d

d/c+c/d=0

This is impossible

a=1/c and b=1/d

d/c+c/d=0

This is impossible

- #11

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Here's what I was hinting at: (ad+bc)^2=0

(ad)^2 + 2abcd + (bc)^2 = 0

(ad)^2 + (bc)^2 = -2

- #12

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d^2+c^2/cd=0

d^2+c^2=0

This is not possible...to be honest I'm not sure how to follow the logic of yours.

- #13

- 64

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d^2+c^2/cd=0

d^2+c^2=0

This is not possible...to be honest I'm not sure how to follow the logic of yours.

What if c and d are 0?

In moving from the second to the third line, I replaced acbd with 1*1.

- #14

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But c and d can't be zero because ac=1 and bd=1

- #15

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But c and d can't be zero because ac=1 and bd=1

Well then, it seems our work here is done.

- #16

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Thanks!

- #17

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

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If x

- #18

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You can use Eisenstein and the fact f(x) is irreducible if and only if f(x+1) is irreducible.

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