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Question about Proof of Real Number..
Q: Prove that for any Real Numbers a,b ab=0⇒a=0 or b=0 Rebecca131 Posts: 3, Reputation: 1 New Member #2 Jun 21, 2017, 12:31 PM
I found the following on Google: "We continue the series of proofs of the basic theorems in Algebra. In this post, we use the Axioms of Real Numbers to prove that the product of any number and zero is zero. You may go back the the preceding link to review the axioms stated below. Theorem For any number a, 0a = 0. Proof We know that 0 + 0 = 0. Multiplying both sides by a, we have (0 + 0)a = 0a. By the Distributive Property (Axiom 4), we get 0a + 0a = 0a. Since by Axiom 5A, any number has an Additive Identity (recall c = c + 0 for any c), we can add 0, the right hand side of the equation giving us 0a + 0a = 0a + 0. By the Cancellation Law, we eliminate one 0a on both sides of the equation resulting to 0a = 0 which is what we want to prove." smoothy Posts: 25,495, Reputation: 2853 Uber Member #3 Jun 21, 2017, 01:16 PM
This is YOUR homework... show us what YOU think the answer is and why. Site rules require it. MathMaven53 Posts: 20, Reputation: 2 New Member #4 Jan 7, 2018, 10:42 AM
a b= 0
If a not equal to 0 we can divide both sides by a
a b divided by a equals b
0 divided by a equals 0
Then b = 0
Similarly assume b not equal to 0
Then divide a b=0 by b

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