-0.000 000 000 742 147 676 536 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 536(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
-0.000 000 000 742 147 676 536(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:

|-0.000 000 000 742 147 676 536| = 0.000 000 000 742 147 676 536


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: 0.000 000 000 742 147 676 536.

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) 0.000 000 000 742 147 676 536 × 2 = 0 + 0.000 000 001 484 295 353 072;
  • 2) 0.000 000 001 484 295 353 072 × 2 = 0 + 0.000 000 002 968 590 706 144;
  • 3) 0.000 000 002 968 590 706 144 × 2 = 0 + 0.000 000 005 937 181 412 288;
  • 4) 0.000 000 005 937 181 412 288 × 2 = 0 + 0.000 000 011 874 362 824 576;
  • 5) 0.000 000 011 874 362 824 576 × 2 = 0 + 0.000 000 023 748 725 649 152;
  • 6) 0.000 000 023 748 725 649 152 × 2 = 0 + 0.000 000 047 497 451 298 304;
  • 7) 0.000 000 047 497 451 298 304 × 2 = 0 + 0.000 000 094 994 902 596 608;
  • 8) 0.000 000 094 994 902 596 608 × 2 = 0 + 0.000 000 189 989 805 193 216;
  • 9) 0.000 000 189 989 805 193 216 × 2 = 0 + 0.000 000 379 979 610 386 432;
  • 10) 0.000 000 379 979 610 386 432 × 2 = 0 + 0.000 000 759 959 220 772 864;
  • 11) 0.000 000 759 959 220 772 864 × 2 = 0 + 0.000 001 519 918 441 545 728;
  • 12) 0.000 001 519 918 441 545 728 × 2 = 0 + 0.000 003 039 836 883 091 456;
  • 13) 0.000 003 039 836 883 091 456 × 2 = 0 + 0.000 006 079 673 766 182 912;
  • 14) 0.000 006 079 673 766 182 912 × 2 = 0 + 0.000 012 159 347 532 365 824;
  • 15) 0.000 012 159 347 532 365 824 × 2 = 0 + 0.000 024 318 695 064 731 648;
  • 16) 0.000 024 318 695 064 731 648 × 2 = 0 + 0.000 048 637 390 129 463 296;
  • 17) 0.000 048 637 390 129 463 296 × 2 = 0 + 0.000 097 274 780 258 926 592;
  • 18) 0.000 097 274 780 258 926 592 × 2 = 0 + 0.000 194 549 560 517 853 184;
  • 19) 0.000 194 549 560 517 853 184 × 2 = 0 + 0.000 389 099 121 035 706 368;
  • 20) 0.000 389 099 121 035 706 368 × 2 = 0 + 0.000 778 198 242 071 412 736;
  • 21) 0.000 778 198 242 071 412 736 × 2 = 0 + 0.001 556 396 484 142 825 472;
  • 22) 0.001 556 396 484 142 825 472 × 2 = 0 + 0.003 112 792 968 285 650 944;
  • 23) 0.003 112 792 968 285 650 944 × 2 = 0 + 0.006 225 585 936 571 301 888;
  • 24) 0.006 225 585 936 571 301 888 × 2 = 0 + 0.012 451 171 873 142 603 776;
  • 25) 0.012 451 171 873 142 603 776 × 2 = 0 + 0.024 902 343 746 285 207 552;
  • 26) 0.024 902 343 746 285 207 552 × 2 = 0 + 0.049 804 687 492 570 415 104;
  • 27) 0.049 804 687 492 570 415 104 × 2 = 0 + 0.099 609 374 985 140 830 208;
  • 28) 0.099 609 374 985 140 830 208 × 2 = 0 + 0.199 218 749 970 281 660 416;
  • 29) 0.199 218 749 970 281 660 416 × 2 = 0 + 0.398 437 499 940 563 320 832;
  • 30) 0.398 437 499 940 563 320 832 × 2 = 0 + 0.796 874 999 881 126 641 664;
  • 31) 0.796 874 999 881 126 641 664 × 2 = 1 + 0.593 749 999 762 253 283 328;
  • 32) 0.593 749 999 762 253 283 328 × 2 = 1 + 0.187 499 999 524 506 566 656;
  • 33) 0.187 499 999 524 506 566 656 × 2 = 0 + 0.374 999 999 049 013 133 312;
  • 34) 0.374 999 999 049 013 133 312 × 2 = 0 + 0.749 999 998 098 026 266 624;
  • 35) 0.749 999 998 098 026 266 624 × 2 = 1 + 0.499 999 996 196 052 533 248;
  • 36) 0.499 999 996 196 052 533 248 × 2 = 0 + 0.999 999 992 392 105 066 496;
  • 37) 0.999 999 992 392 105 066 496 × 2 = 1 + 0.999 999 984 784 210 132 992;
  • 38) 0.999 999 984 784 210 132 992 × 2 = 1 + 0.999 999 969 568 420 265 984;
  • 39) 0.999 999 969 568 420 265 984 × 2 = 1 + 0.999 999 939 136 840 531 968;
  • 40) 0.999 999 939 136 840 531 968 × 2 = 1 + 0.999 999 878 273 681 063 936;
  • 41) 0.999 999 878 273 681 063 936 × 2 = 1 + 0.999 999 756 547 362 127 872;
  • 42) 0.999 999 756 547 362 127 872 × 2 = 1 + 0.999 999 513 094 724 255 744;
  • 43) 0.999 999 513 094 724 255 744 × 2 = 1 + 0.999 999 026 189 448 511 488;
  • 44) 0.999 999 026 189 448 511 488 × 2 = 1 + 0.999 998 052 378 897 022 976;
  • 45) 0.999 998 052 378 897 022 976 × 2 = 1 + 0.999 996 104 757 794 045 952;
  • 46) 0.999 996 104 757 794 045 952 × 2 = 1 + 0.999 992 209 515 588 091 904;
  • 47) 0.999 992 209 515 588 091 904 × 2 = 1 + 0.999 984 419 031 176 183 808;
  • 48) 0.999 984 419 031 176 183 808 × 2 = 1 + 0.999 968 838 062 352 367 616;
  • 49) 0.999 968 838 062 352 367 616 × 2 = 1 + 0.999 937 676 124 704 735 232;
  • 50) 0.999 937 676 124 704 735 232 × 2 = 1 + 0.999 875 352 249 409 470 464;
  • 51) 0.999 875 352 249 409 470 464 × 2 = 1 + 0.999 750 704 498 818 940 928;
  • 52) 0.999 750 704 498 818 940 928 × 2 = 1 + 0.999 501 408 997 637 881 856;
  • 53) 0.999 501 408 997 637 881 856 × 2 = 1 + 0.999 002 817 995 275 763 712;
  • 54) 0.999 002 817 995 275 763 712 × 2 = 1 + 0.998 005 635 990 551 527 424;

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:


0.000 000 000 742 147 676 536(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 536(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:


0.000 000 000 742 147 676 536(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 -0.000 000 000 742 147 676 536 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