-0.000 000 000 742 147 676 646 543 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 543(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
-0.000 000 000 742 147 676 646 543(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:

|-0.000 000 000 742 147 676 646 543| = 0.000 000 000 742 147 676 646 543


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: 0.000 000 000 742 147 676 646 543.

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) 0.000 000 000 742 147 676 646 543 × 2 = 0 + 0.000 000 001 484 295 353 293 086;
  • 2) 0.000 000 001 484 295 353 293 086 × 2 = 0 + 0.000 000 002 968 590 706 586 172;
  • 3) 0.000 000 002 968 590 706 586 172 × 2 = 0 + 0.000 000 005 937 181 413 172 344;
  • 4) 0.000 000 005 937 181 413 172 344 × 2 = 0 + 0.000 000 011 874 362 826 344 688;
  • 5) 0.000 000 011 874 362 826 344 688 × 2 = 0 + 0.000 000 023 748 725 652 689 376;
  • 6) 0.000 000 023 748 725 652 689 376 × 2 = 0 + 0.000 000 047 497 451 305 378 752;
  • 7) 0.000 000 047 497 451 305 378 752 × 2 = 0 + 0.000 000 094 994 902 610 757 504;
  • 8) 0.000 000 094 994 902 610 757 504 × 2 = 0 + 0.000 000 189 989 805 221 515 008;
  • 9) 0.000 000 189 989 805 221 515 008 × 2 = 0 + 0.000 000 379 979 610 443 030 016;
  • 10) 0.000 000 379 979 610 443 030 016 × 2 = 0 + 0.000 000 759 959 220 886 060 032;
  • 11) 0.000 000 759 959 220 886 060 032 × 2 = 0 + 0.000 001 519 918 441 772 120 064;
  • 12) 0.000 001 519 918 441 772 120 064 × 2 = 0 + 0.000 003 039 836 883 544 240 128;
  • 13) 0.000 003 039 836 883 544 240 128 × 2 = 0 + 0.000 006 079 673 767 088 480 256;
  • 14) 0.000 006 079 673 767 088 480 256 × 2 = 0 + 0.000 012 159 347 534 176 960 512;
  • 15) 0.000 012 159 347 534 176 960 512 × 2 = 0 + 0.000 024 318 695 068 353 921 024;
  • 16) 0.000 024 318 695 068 353 921 024 × 2 = 0 + 0.000 048 637 390 136 707 842 048;
  • 17) 0.000 048 637 390 136 707 842 048 × 2 = 0 + 0.000 097 274 780 273 415 684 096;
  • 18) 0.000 097 274 780 273 415 684 096 × 2 = 0 + 0.000 194 549 560 546 831 368 192;
  • 19) 0.000 194 549 560 546 831 368 192 × 2 = 0 + 0.000 389 099 121 093 662 736 384;
  • 20) 0.000 389 099 121 093 662 736 384 × 2 = 0 + 0.000 778 198 242 187 325 472 768;
  • 21) 0.000 778 198 242 187 325 472 768 × 2 = 0 + 0.001 556 396 484 374 650 945 536;
  • 22) 0.001 556 396 484 374 650 945 536 × 2 = 0 + 0.003 112 792 968 749 301 891 072;
  • 23) 0.003 112 792 968 749 301 891 072 × 2 = 0 + 0.006 225 585 937 498 603 782 144;
  • 24) 0.006 225 585 937 498 603 782 144 × 2 = 0 + 0.012 451 171 874 997 207 564 288;
  • 25) 0.012 451 171 874 997 207 564 288 × 2 = 0 + 0.024 902 343 749 994 415 128 576;
  • 26) 0.024 902 343 749 994 415 128 576 × 2 = 0 + 0.049 804 687 499 988 830 257 152;
  • 27) 0.049 804 687 499 988 830 257 152 × 2 = 0 + 0.099 609 374 999 977 660 514 304;
  • 28) 0.099 609 374 999 977 660 514 304 × 2 = 0 + 0.199 218 749 999 955 321 028 608;
  • 29) 0.199 218 749 999 955 321 028 608 × 2 = 0 + 0.398 437 499 999 910 642 057 216;
  • 30) 0.398 437 499 999 910 642 057 216 × 2 = 0 + 0.796 874 999 999 821 284 114 432;
  • 31) 0.796 874 999 999 821 284 114 432 × 2 = 1 + 0.593 749 999 999 642 568 228 864;
  • 32) 0.593 749 999 999 642 568 228 864 × 2 = 1 + 0.187 499 999 999 285 136 457 728;
  • 33) 0.187 499 999 999 285 136 457 728 × 2 = 0 + 0.374 999 999 998 570 272 915 456;
  • 34) 0.374 999 999 998 570 272 915 456 × 2 = 0 + 0.749 999 999 997 140 545 830 912;
  • 35) 0.749 999 999 997 140 545 830 912 × 2 = 1 + 0.499 999 999 994 281 091 661 824;
  • 36) 0.499 999 999 994 281 091 661 824 × 2 = 0 + 0.999 999 999 988 562 183 323 648;
  • 37) 0.999 999 999 988 562 183 323 648 × 2 = 1 + 0.999 999 999 977 124 366 647 296;
  • 38) 0.999 999 999 977 124 366 647 296 × 2 = 1 + 0.999 999 999 954 248 733 294 592;
  • 39) 0.999 999 999 954 248 733 294 592 × 2 = 1 + 0.999 999 999 908 497 466 589 184;
  • 40) 0.999 999 999 908 497 466 589 184 × 2 = 1 + 0.999 999 999 816 994 933 178 368;
  • 41) 0.999 999 999 816 994 933 178 368 × 2 = 1 + 0.999 999 999 633 989 866 356 736;
  • 42) 0.999 999 999 633 989 866 356 736 × 2 = 1 + 0.999 999 999 267 979 732 713 472;
  • 43) 0.999 999 999 267 979 732 713 472 × 2 = 1 + 0.999 999 998 535 959 465 426 944;
  • 44) 0.999 999 998 535 959 465 426 944 × 2 = 1 + 0.999 999 997 071 918 930 853 888;
  • 45) 0.999 999 997 071 918 930 853 888 × 2 = 1 + 0.999 999 994 143 837 861 707 776;
  • 46) 0.999 999 994 143 837 861 707 776 × 2 = 1 + 0.999 999 988 287 675 723 415 552;
  • 47) 0.999 999 988 287 675 723 415 552 × 2 = 1 + 0.999 999 976 575 351 446 831 104;
  • 48) 0.999 999 976 575 351 446 831 104 × 2 = 1 + 0.999 999 953 150 702 893 662 208;
  • 49) 0.999 999 953 150 702 893 662 208 × 2 = 1 + 0.999 999 906 301 405 787 324 416;
  • 50) 0.999 999 906 301 405 787 324 416 × 2 = 1 + 0.999 999 812 602 811 574 648 832;
  • 51) 0.999 999 812 602 811 574 648 832 × 2 = 1 + 0.999 999 625 205 623 149 297 664;
  • 52) 0.999 999 625 205 623 149 297 664 × 2 = 1 + 0.999 999 250 411 246 298 595 328;
  • 53) 0.999 999 250 411 246 298 595 328 × 2 = 1 + 0.999 998 500 822 492 597 190 656;
  • 54) 0.999 998 500 822 492 597 190 656 × 2 = 1 + 0.999 997 001 644 985 194 381 312;

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:


0.000 000 000 742 147 676 646 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 543(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 543 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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