Answer to Question #106146 in Real Analysis for lessinah

Question #106146
Use a proof bt of the set of n1 and n2 is less than n^1/2y contradiction to establish the following:
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2
1
Expert's answer
2020-03-24T13:22:06-0400

n=n1n2,n=n_1n_2, n12,n22n_1\geq2 , n_2\geq2 .

Use a proof by contradiction. Let us assume that both of n1,n2n_1,\,n_2 are greater than or equal to n1/2.n^{1/2}.

Case I

n1=n1/2n_1=n^{1/2}

n2=n1/2n_2=n^{1/2}

n1n2=n12+12=nn_1n_2=n^{\frac12+\frac12}=n

Case II

n1=n12+x;x>0n_1=n^{\frac12+x}; x>0

n2=n12+y;y>0n_2=n^{\frac12+y}; y>0

n1n2=n12+xn12+y;x>0;y>0n_1n_2=n^{\frac12+x}n^{\frac12+y}; x>0 ; y>0

n1n2=nnx+y;x>0;y>0n_1n_2=nn^{x+y}; x>0 ; y>0

So, x+y=0x+y=0

x=yx=-y but x>0,y>0x>0,y>0

which is a contradiction.

So,If a positive whole number nn can be expressed as n1n2n_1 n_2 , where n1n_1 is greater than or equal to 2 and n2n_2 is greater than or equal to 2, then at least one element of n1n_1 and n2n_2 is less than n1/2n^{1/2} .



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment