Proof 1 Suppose 1 = 2. Then we can double both sides and get 2 = 4. We can then

Question

Proof 1

Suppose 1 = 2. Then we can double both sides and get 2 = 4. We can then subtract three from both sides and get -1 = 1. Finally

we can square both sides giving that 1 = 1. This is true so 1 = 2.

Proof 2

Let a,x be real numbers with a = x. We can then add a to both sides giving 2a = a+x. We can then subtract 2x from both sides

giving 2a-2x=a-x. Factoring 2 out of the left sides gives 2(a-x) = a-x. We can then cancel a-x from both sides and get 2=1.

Therefore1=2.

Proof 3

Let x = 0. Then x = 0+0+0+... = (1+(-1))+(1+(-1))+(1+(-1))+.... Addition is associative so we can remove the parenthases safely.

Thus, x = 1+(-1)+1+(-1)+1+(-1)+.... The number of terms that are shown won't alter the sum so x=1+(-1)+1+(-1)+1+.... We can

reinsert parenthases and getx=1+((-1)+1)+((-1)+1)+...=1+0+0+.... Thus,x=1. Therefore 0=1.

Proof 4

Letx=n. Then x=1+...+1(n times). Multiplying both sides by x gives x =x+...+x(n times). Taking a derivative of

both sides gives 2x=1+...+1(n times)=n. Then 2x=n=x so 2=1.

Explanation / Answer

proof 1:

if a^2 = b^2 then it may not be true that a = b. We cannot conclude that.

proof 2:

you cannot cancel (a-x) on both sides, because its value is 0. you cannot divide by 0.

proof3:

the number sequence of x = (1+(-1))+(1+(-1))+(1+(-1))+.... has to end somewhere, so x = 1+((-1)+1)+((-1)+1)+... has a -1 term at the end. so its value is 0 not 1.

proof4:

when derivative is taken, we cannot take n as constant as it is dependent on x. that is what is wrong with the proof.

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