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# proof by contraposition watch

1. so then

x by (x²+y²)>0

then the usual -q->-p so the p->q

but i don't understand how you suddenly get "x by (x²+y²)>0" i don't get where it comes from and where the following statement comes from either it makes no sense to me, can someone kindly explain please?
2. (Original post by will'o'wisp2)

so then

x by (x²+y²)>0

then the usual -q->-p so the p->q

but i don't understand how you suddenly get "x by (x²+y²)>0" i don't get where it comes from and where the following statement comes from either it makes no sense to me, can someone kindly explain please?
I'm struggling to follow your notation a bit, but I think what you've called "-q" is basically "Not q" in logical notation.

So you're assuming that statement Not q is true, that is, that y > x.

We can multiply both sides of this inequality by any positive number without affecting the direction of the inequality, and an example of a positive number (plucked out of thin air a bit!) is x^2 + y^2.

So starting from y > x we multiply by x^2 + y^2 to get

y(x^2 + y^2) > x(x^2 + y^2)

If you expand the brackets on both sides, you end up with the opposite inequality to the one you were first given, i.e. by assuming "Not q" you have proved "Not p".

From this, you can infer that p implies q.

This is assuming I've followed your notation correctly
3. (Original post by davros)
I'm struggling to follow your notation a bit, but I think what you've called "-q" is basically "Not q" in logical notation.

So you're assuming that statement Not q is true, that is, that y > x.

We can multiply both sides of this inequality by any positive number without affecting the direction of the inequality, and an example of a positive number (plucked out of thin air a bit!) is x^2 + y^2.

So starting from y > x we multiply by x^2 + y^2 to get

y(x^2 + y^2) > x(x^2 + y^2)

If you expand the brackets on both sides, you end up with the opposite inequality to the one you were first given, i.e. by assuming "Not q" you have proved "Not p".

From this, you can infer that p implies q.

This is assuming I've followed your notation correctly
tthanks you so much

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Updated: October 15, 2017
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