1) p is always less than or equal to minimum of x,y,z.
2)p is alwasys greater than or equal to maximum of x,y,z.
3) p is always greater than the minimum of x,y,z and less than the maximum of x,y,z.
4) none of these.
2007-06-13 02:30:35 · 5 answers · asked by Targeting C.A.T 1 in Science & Mathematics ➔ Mathematics
Try some examples you know and at least see if you can cancel out any possibilities.
3^2 + 4^2 = 5^2
This rules out 2 since p is NOT greater than 5. Choice 3 is also out since 2 is not between 3 and 5.
What about choice 1? Hmmm...
Fermat's Last Theorem:
If an integer n is greater than 2, then the equation a^n + b^n = c^n has no solutions in non-zero integers a, b, and c.
So IF x^p+y^p = z^p then either...
p = 2
x = 0
y = 0
Of course 0 isn't a natural number, so that leaves us with p=2. The smallest values of x,y and z that work are the ever-popular 3,4 and 5.
So x,y and z are ALWAYS greater than p. Choice 1.
2007-06-13 03:01:59 · answer #1 · answered by ryanker1 4 · 1⤊ 0⤋
2
2007-06-13 02:40:45 · answer #2 · answered by Nik 2 · 0⤊ 1⤋
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2016-10-07 10:23:30 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Ans 1)
Pf: Let x be smallest and set
x^p = (z-y)(z^(p-1) + z^(p-2)*y +.....+z*y^(p-2) + y^(p-1) )
put x in formula and since x<=y and (z-y)>=1 we have
x^p >= 1( p*x^(p-1) ) . Dividing by x^(p-1) yields , x>=p.
2007-06-13 02:56:18 · answer #4 · answered by knashha 5 · 2⤊ 0⤋
p<=2 so 4)none of these (Fermat´s last theorem)
2007-06-13 07:17:43 · answer #5 · answered by santmann2002 7 · 0⤊ 0⤋