-0.000 000 000 000 000 000 000 083 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 000 083(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 000 000 083(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 000 000 083| = 0.000 000 000 000 000 000 000 083


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 000 083.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 000 000 083 × 2 = 0 + 0.000 000 000 000 000 000 000 166;
  • 2) 0.000 000 000 000 000 000 000 166 × 2 = 0 + 0.000 000 000 000 000 000 000 332;
  • 3) 0.000 000 000 000 000 000 000 332 × 2 = 0 + 0.000 000 000 000 000 000 000 664;
  • 4) 0.000 000 000 000 000 000 000 664 × 2 = 0 + 0.000 000 000 000 000 000 001 328;
  • 5) 0.000 000 000 000 000 000 001 328 × 2 = 0 + 0.000 000 000 000 000 000 002 656;
  • 6) 0.000 000 000 000 000 000 002 656 × 2 = 0 + 0.000 000 000 000 000 000 005 312;
  • 7) 0.000 000 000 000 000 000 005 312 × 2 = 0 + 0.000 000 000 000 000 000 010 624;
  • 8) 0.000 000 000 000 000 000 010 624 × 2 = 0 + 0.000 000 000 000 000 000 021 248;
  • 9) 0.000 000 000 000 000 000 021 248 × 2 = 0 + 0.000 000 000 000 000 000 042 496;
  • 10) 0.000 000 000 000 000 000 042 496 × 2 = 0 + 0.000 000 000 000 000 000 084 992;
  • 11) 0.000 000 000 000 000 000 084 992 × 2 = 0 + 0.000 000 000 000 000 000 169 984;
  • 12) 0.000 000 000 000 000 000 169 984 × 2 = 0 + 0.000 000 000 000 000 000 339 968;
  • 13) 0.000 000 000 000 000 000 339 968 × 2 = 0 + 0.000 000 000 000 000 000 679 936;
  • 14) 0.000 000 000 000 000 000 679 936 × 2 = 0 + 0.000 000 000 000 000 001 359 872;
  • 15) 0.000 000 000 000 000 001 359 872 × 2 = 0 + 0.000 000 000 000 000 002 719 744;
  • 16) 0.000 000 000 000 000 002 719 744 × 2 = 0 + 0.000 000 000 000 000 005 439 488;
  • 17) 0.000 000 000 000 000 005 439 488 × 2 = 0 + 0.000 000 000 000 000 010 878 976;
  • 18) 0.000 000 000 000 000 010 878 976 × 2 = 0 + 0.000 000 000 000 000 021 757 952;
  • 19) 0.000 000 000 000 000 021 757 952 × 2 = 0 + 0.000 000 000 000 000 043 515 904;
  • 20) 0.000 000 000 000 000 043 515 904 × 2 = 0 + 0.000 000 000 000 000 087 031 808;
  • 21) 0.000 000 000 000 000 087 031 808 × 2 = 0 + 0.000 000 000 000 000 174 063 616;
  • 22) 0.000 000 000 000 000 174 063 616 × 2 = 0 + 0.000 000 000 000 000 348 127 232;
  • 23) 0.000 000 000 000 000 348 127 232 × 2 = 0 + 0.000 000 000 000 000 696 254 464;
  • 24) 0.000 000 000 000 000 696 254 464 × 2 = 0 + 0.000 000 000 000 001 392 508 928;
  • 25) 0.000 000 000 000 001 392 508 928 × 2 = 0 + 0.000 000 000 000 002 785 017 856;
  • 26) 0.000 000 000 000 002 785 017 856 × 2 = 0 + 0.000 000 000 000 005 570 035 712;
  • 27) 0.000 000 000 000 005 570 035 712 × 2 = 0 + 0.000 000 000 000 011 140 071 424;
  • 28) 0.000 000 000 000 011 140 071 424 × 2 = 0 + 0.000 000 000 000 022 280 142 848;
  • 29) 0.000 000 000 000 022 280 142 848 × 2 = 0 + 0.000 000 000 000 044 560 285 696;
  • 30) 0.000 000 000 000 044 560 285 696 × 2 = 0 + 0.000 000 000 000 089 120 571 392;
  • 31) 0.000 000 000 000 089 120 571 392 × 2 = 0 + 0.000 000 000 000 178 241 142 784;
  • 32) 0.000 000 000 000 178 241 142 784 × 2 = 0 + 0.000 000 000 000 356 482 285 568;
  • 33) 0.000 000 000 000 356 482 285 568 × 2 = 0 + 0.000 000 000 000 712 964 571 136;
  • 34) 0.000 000 000 000 712 964 571 136 × 2 = 0 + 0.000 000 000 001 425 929 142 272;
  • 35) 0.000 000 000 001 425 929 142 272 × 2 = 0 + 0.000 000 000 002 851 858 284 544;
  • 36) 0.000 000 000 002 851 858 284 544 × 2 = 0 + 0.000 000 000 005 703 716 569 088;
  • 37) 0.000 000 000 005 703 716 569 088 × 2 = 0 + 0.000 000 000 011 407 433 138 176;
  • 38) 0.000 000 000 011 407 433 138 176 × 2 = 0 + 0.000 000 000 022 814 866 276 352;
  • 39) 0.000 000 000 022 814 866 276 352 × 2 = 0 + 0.000 000 000 045 629 732 552 704;
  • 40) 0.000 000 000 045 629 732 552 704 × 2 = 0 + 0.000 000 000 091 259 465 105 408;
  • 41) 0.000 000 000 091 259 465 105 408 × 2 = 0 + 0.000 000 000 182 518 930 210 816;
  • 42) 0.000 000 000 182 518 930 210 816 × 2 = 0 + 0.000 000 000 365 037 860 421 632;
  • 43) 0.000 000 000 365 037 860 421 632 × 2 = 0 + 0.000 000 000 730 075 720 843 264;
  • 44) 0.000 000 000 730 075 720 843 264 × 2 = 0 + 0.000 000 001 460 151 441 686 528;
  • 45) 0.000 000 001 460 151 441 686 528 × 2 = 0 + 0.000 000 002 920 302 883 373 056;
  • 46) 0.000 000 002 920 302 883 373 056 × 2 = 0 + 0.000 000 005 840 605 766 746 112;
  • 47) 0.000 000 005 840 605 766 746 112 × 2 = 0 + 0.000 000 011 681 211 533 492 224;
  • 48) 0.000 000 011 681 211 533 492 224 × 2 = 0 + 0.000 000 023 362 423 066 984 448;
  • 49) 0.000 000 023 362 423 066 984 448 × 2 = 0 + 0.000 000 046 724 846 133 968 896;
  • 50) 0.000 000 046 724 846 133 968 896 × 2 = 0 + 0.000 000 093 449 692 267 937 792;
  • 51) 0.000 000 093 449 692 267 937 792 × 2 = 0 + 0.000 000 186 899 384 535 875 584;
  • 52) 0.000 000 186 899 384 535 875 584 × 2 = 0 + 0.000 000 373 798 769 071 751 168;
  • 53) 0.000 000 373 798 769 071 751 168 × 2 = 0 + 0.000 000 747 597 538 143 502 336;
  • 54) 0.000 000 747 597 538 143 502 336 × 2 = 0 + 0.000 001 495 195 076 287 004 672;
  • 55) 0.000 001 495 195 076 287 004 672 × 2 = 0 + 0.000 002 990 390 152 574 009 344;
  • 56) 0.000 002 990 390 152 574 009 344 × 2 = 0 + 0.000 005 980 780 305 148 018 688;
  • 57) 0.000 005 980 780 305 148 018 688 × 2 = 0 + 0.000 011 961 560 610 296 037 376;
  • 58) 0.000 011 961 560 610 296 037 376 × 2 = 0 + 0.000 023 923 121 220 592 074 752;
  • 59) 0.000 023 923 121 220 592 074 752 × 2 = 0 + 0.000 047 846 242 441 184 149 504;
  • 60) 0.000 047 846 242 441 184 149 504 × 2 = 0 + 0.000 095 692 484 882 368 299 008;
  • 61) 0.000 095 692 484 882 368 299 008 × 2 = 0 + 0.000 191 384 969 764 736 598 016;
  • 62) 0.000 191 384 969 764 736 598 016 × 2 = 0 + 0.000 382 769 939 529 473 196 032;
  • 63) 0.000 382 769 939 529 473 196 032 × 2 = 0 + 0.000 765 539 879 058 946 392 064;
  • 64) 0.000 765 539 879 058 946 392 064 × 2 = 0 + 0.001 531 079 758 117 892 784 128;
  • 65) 0.001 531 079 758 117 892 784 128 × 2 = 0 + 0.003 062 159 516 235 785 568 256;
  • 66) 0.003 062 159 516 235 785 568 256 × 2 = 0 + 0.006 124 319 032 471 571 136 512;
  • 67) 0.006 124 319 032 471 571 136 512 × 2 = 0 + 0.012 248 638 064 943 142 273 024;
  • 68) 0.012 248 638 064 943 142 273 024 × 2 = 0 + 0.024 497 276 129 886 284 546 048;
  • 69) 0.024 497 276 129 886 284 546 048 × 2 = 0 + 0.048 994 552 259 772 569 092 096;
  • 70) 0.048 994 552 259 772 569 092 096 × 2 = 0 + 0.097 989 104 519 545 138 184 192;
  • 71) 0.097 989 104 519 545 138 184 192 × 2 = 0 + 0.195 978 209 039 090 276 368 384;
  • 72) 0.195 978 209 039 090 276 368 384 × 2 = 0 + 0.391 956 418 078 180 552 736 768;
  • 73) 0.391 956 418 078 180 552 736 768 × 2 = 0 + 0.783 912 836 156 361 105 473 536;
  • 74) 0.783 912 836 156 361 105 473 536 × 2 = 1 + 0.567 825 672 312 722 210 947 072;
  • 75) 0.567 825 672 312 722 210 947 072 × 2 = 1 + 0.135 651 344 625 444 421 894 144;
  • 76) 0.135 651 344 625 444 421 894 144 × 2 = 0 + 0.271 302 689 250 888 843 788 288;
  • 77) 0.271 302 689 250 888 843 788 288 × 2 = 0 + 0.542 605 378 501 777 687 576 576;
  • 78) 0.542 605 378 501 777 687 576 576 × 2 = 1 + 0.085 210 757 003 555 375 153 152;
  • 79) 0.085 210 757 003 555 375 153 152 × 2 = 0 + 0.170 421 514 007 110 750 306 304;
  • 80) 0.170 421 514 007 110 750 306 304 × 2 = 0 + 0.340 843 028 014 221 500 612 608;
  • 81) 0.340 843 028 014 221 500 612 608 × 2 = 0 + 0.681 686 056 028 443 001 225 216;
  • 82) 0.681 686 056 028 443 001 225 216 × 2 = 1 + 0.363 372 112 056 886 002 450 432;
  • 83) 0.363 372 112 056 886 002 450 432 × 2 = 0 + 0.726 744 224 113 772 004 900 864;
  • 84) 0.726 744 224 113 772 004 900 864 × 2 = 1 + 0.453 488 448 227 544 009 801 728;
  • 85) 0.453 488 448 227 544 009 801 728 × 2 = 0 + 0.906 976 896 455 088 019 603 456;
  • 86) 0.906 976 896 455 088 019 603 456 × 2 = 1 + 0.813 953 792 910 176 039 206 912;
  • 87) 0.813 953 792 910 176 039 206 912 × 2 = 1 + 0.627 907 585 820 352 078 413 824;
  • 88) 0.627 907 585 820 352 078 413 824 × 2 = 1 + 0.255 815 171 640 704 156 827 648;
  • 89) 0.255 815 171 640 704 156 827 648 × 2 = 0 + 0.511 630 343 281 408 313 655 296;
  • 90) 0.511 630 343 281 408 313 655 296 × 2 = 1 + 0.023 260 686 562 816 627 310 592;
  • 91) 0.023 260 686 562 816 627 310 592 × 2 = 0 + 0.046 521 373 125 633 254 621 184;
  • 92) 0.046 521 373 125 633 254 621 184 × 2 = 0 + 0.093 042 746 251 266 509 242 368;
  • 93) 0.093 042 746 251 266 509 242 368 × 2 = 0 + 0.186 085 492 502 533 018 484 736;
  • 94) 0.186 085 492 502 533 018 484 736 × 2 = 0 + 0.372 170 985 005 066 036 969 472;
  • 95) 0.372 170 985 005 066 036 969 472 × 2 = 0 + 0.744 341 970 010 132 073 938 944;
  • 96) 0.744 341 970 010 132 073 938 944 × 2 = 1 + 0.488 683 940 020 264 147 877 888;
  • 97) 0.488 683 940 020 264 147 877 888 × 2 = 0 + 0.977 367 880 040 528 295 755 776;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 000 000 083(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0100 0101 0111 0100 0001 0(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 000 083(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0100 0101 0111 0100 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 74 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 000 083(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0100 0101 0111 0100 0001 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0100 0101 0111 0100 0001 0(2) × 20 =

1.1001 0001 0101 1101 0000 010(2) × 2-74


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -74

Mantissa (not normalized):
1.1001 0001 0101 1101 0000 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-74 + 2(8-1) - 1 =

(-74 + 127)(10) =

53(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

53(10) =

0011 0101(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1000 1010 1110 1000 0010 =

100 1000 1010 1110 1000 0010


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0011 0101

Mantissa (23 bits) =
100 1000 1010 1110 1000 0010


Decimal number -0.000 000 000 000 000 000 000 083 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0011 0101 - 100 1000 1010 1110 1000 0010

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111