Question

2. Determine if the proposition below is true or false. Justify all your conclusions. If a biconditional statement is found to be false, you should clearly determine if one of the conditional statements within it is true. In that case, you should state an appropriate theorem for this conditional statement and prove it. Proposition 1. For all integers m and n, 4 divides (m2 – n) if and only if m and n are both even or m and n are both odd.

Answer 1

Tool-d if ne is Even then n is Even I proof By contrapositive. r 6=) q is Equivalent O to 79-7p NOT Q → Not el je "lf in is not even then ne is not even. " le If n is odd then ne is Odd KGZI n= 2K+1 n2 = (2K413² = UK² +4K tl = 2( 262 +2K)+ - n2 = ak'ti KEZ so when n is odd nz is odd. so NOT a = not p as so na Even is Even proved by then n is contra positive

Tool. 2 ir ne is odd then n is odd. I proof By contrapositive. is not odd "If n then ² is Not oddi 2) "if n is Even then 22 is Even" a na ak. KEZI a p2 = 2(2K2) n2 is odd then n is odd. proved By contrapositive po contrapositive NOT Q = NOT P Now COMING TO OUR QUESTION proposition for all integer m and n 4 Divides m² n² lff m and n are both even or both odd. TRUE

Proof converse pard Fast Ik mand n are = maakt Both even odd . ,n=ak't for some Kokeze + m2-n2 = (2x+132 - (2k! 4133 = (UK²+UK+1)-(4x12 +4K'+1) 1. = 4K²_ukl2 +4K. 4k tly = u(K?+K –K2_K") Rik'tz K+R 442 kl = KX EZ a K*Er. m² n2 = uka u/m²_n2 a If & m and n are Both Even a m=2k , n=2k' Kakit te m2_n2 = UK2_4k12. - uCK?K12) 41 m2_n a

c m and and n Both m Even on If Both so - odd then a m?na. divides NOW If ust part u1m² n2 then prove 1 that Even or Both Either Both Odd. m and n proof proof By contradiction Suppose if possible is m is Even m and n is odd llig m is we will prove Cases NEVER so they their is Both m and odd. I odd and n is even that Above two HAPPEN (By contradiction) only case Left are n are Even OR Both - then odd i m is Even and n is m= 2K na 2k'+1 m2-n2 = UK2- (2x'+792 m²-n2 = uk² _ UK12_1-4k &

3 m2-n2 = 4(K2_K2_4k') tes y oo uke!) top | contradiction to ulm-n? But up (K2_k12 so this is a Our Assumption so m cant be I can't be odd n Even and together Similarly (2 m and n is Even is odd me 2R+ na ZKI m²_n2 = (2K+172 - (2,132 = UK²+uk-ukla ti o u/m2 na in-guen . But af ulk?ik_k12)+1 so contradiction to our Assumption so m. cant be odd and n cant I be Even fogether. so By Both the cases proved that L a

- m and NA n should Both odd. be Both even - or HENCE PROVED Another Proof (Extrag) Proof 2 for partid Let u/m²_n2 s m²-n2 = uk a m? = UK +12 Now if n is odd as m2 = UK + (2x1133 = UK + 4K!? + K +1 - m2 = 42(2x+ 24 +212) +1 - m2 = 2k²+1 m4 Even) I'm2 is odd a m is odd By Tool-2 s m is odd so In odd a m is odd. I het of n is Even mze UK + (2x) = 4K + 4 =2(214241 mais Even om is Even so in Even am Even Similar proef m odd for yn odd By im Even a n Even Both Tool so both m or Even and n Both are odd, Hence provet

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