m2 chap.0 function

I can help to answer your Q.

(a) Put both a = b = 1

f( 1 x 1 ) = f( 1 ) + f ( 1 )

f( 1 ) = 2 f( 1 )

f( 1 ) = 0
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Put a = -1 and b = -1

f( -1 x -1 ) = f( -1 ) + f ( -1 )

f( 1 ) = 2 f( -1 )

0 = 2 f( -1 )

f( -1 ) = 0
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(b) Show f( -ab ) = f( ab ) for all real a and b.

f( -ab ) = f( a ) + f( -b )

f( -ab ) = f( a ) + f( -1b )

f( -ab ) = f( a ) + f( -1 ) + f( b )

f( -ab ) = f( a ) + f( b ) + f( -1 )

f( -ab ) = f( ab ) + 0

f( -ab ) = f( ab )

Hence, f is even
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Hope I can help you.

(a) f(1*1)=f(1)+f(1)
f(1)=f(1)+f(1)
f(1)=0

f(1*-1)=f(1)+f(-1)
f(-1)=0+f(-1)
f(-1)=0

(b) Now f(-1*x)=f(-1)+f(x)
f(-x)=f(x) [by the result of (a)]
So, y=f(x) is an even function

2011-01-10 23:34:50 補充：
f(1)=f(1)+f(1)
f(1)-f(1)=f(1)+f(1)-f(1)
0=f(1)