Given that $$a^{2}\le a^{2}+b^{2},$$ how can I prove that $$\sqrt{a^{2}}\le \sqrt{a^{2}+b^{2}}$$ strictly from the definition of an ordered field? A derivation step by step would be appreciated.
2026-03-26 14:42:49.1774536169
How can I prove $\sqrt{a^{2}}\le \sqrt{a^{2}+b^{2}}$ if $a^{2}\le a^{2}+b^{2}$?
125 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in REAL-ANALYSIS
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