need help with the proofs for allSolutionIf possible let x26

need help with the proofs for all

If possible let x^2=6 has a solution m = p/q in its lowest form

p^2/q^2 = 6

p^2 = 6q^2

Since p^2 is a perfect square q^2 must have a factor 6.

Or q must have a factor 6.

Then p^2 = 36r^2 or p will have a factor of 6.

Contradiction to the fact p/q is in its lowest form

Hence x^2 =6 cannot have a rational solution

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Let a*b =0

and if possible a and b not equal to 0

Then since a is non zero there is a real number 1/a such that a*1/a = 1

Similarly b*1/b =1

Or ab (1/ab) = 1

But ab =0

So this is not possible.

Hence our assumption was wrong

Either a or b or both should be 0

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Given that a<b

i.e. there is a h >0 such that

a+h = b

Add c to both sides

a+c+h = b+c

or a+c>b+c