# C1 - Prove that...Watch

#1
7b) Given that x and y are integers, prove that

1/(x - "y) + 1/(x + "y)

can be written in the form p/q where p and q are both intergers.

(" is the square root of)

Hope someone can help!!
0
13 years ago
#2
(Original post by lapsaj)
7b) Given that x and y are integers, prove that

1/(x - "y) + 1/(x + "y)

can be written in the form p/q where p and q are both intergers.
1/(x-√y) + 1/(x+√y) Put over common denominator of (x-√y)(x+√y)
= (x+√y)/[(x-√y)(x+√y)] + (x-√y)/[(x-√y)(x+√y)]
But (x-√y)(x+√y) = xÂ²-y

So insert that in to get:
(x+√y+x-√y)/(xÂ²-y)
= 2x/(xÂ²-y)

Since x and y are integers, it can be written as p/q
0
13 years ago
#3
1/(x - √y) + 1/(x + √y) = (x+√y + x - √y)/(x+√y)(x-√y) = 2x/(x^2 - y)

both numerator and rational, given x,y integers. so it's rational.
0
13 years ago
#4
If you simplify the expression:

[1 / (x - √y)] + [1 / (x + √y)]
= [(x + √y) + (x - √y)] / (x + √y)(x - √y)
= 2x / (xÂ² - y)

Since x and y are both integers, 2x will be an integer value, and xÂ² - y will also be an integer value.
0
#5
Thanks a lot guys, amazed at the really quick response!!
0
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