-0.000 000 000 742 147 675 928 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 928(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
-0.000 000 000 742 147 675 928(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:

|-0.000 000 000 742 147 675 928| = 0.000 000 000 742 147 675 928


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: 0.000 000 000 742 147 675 928.

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) 0.000 000 000 742 147 675 928 × 2 = 0 + 0.000 000 001 484 295 351 856;
  • 2) 0.000 000 001 484 295 351 856 × 2 = 0 + 0.000 000 002 968 590 703 712;
  • 3) 0.000 000 002 968 590 703 712 × 2 = 0 + 0.000 000 005 937 181 407 424;
  • 4) 0.000 000 005 937 181 407 424 × 2 = 0 + 0.000 000 011 874 362 814 848;
  • 5) 0.000 000 011 874 362 814 848 × 2 = 0 + 0.000 000 023 748 725 629 696;
  • 6) 0.000 000 023 748 725 629 696 × 2 = 0 + 0.000 000 047 497 451 259 392;
  • 7) 0.000 000 047 497 451 259 392 × 2 = 0 + 0.000 000 094 994 902 518 784;
  • 8) 0.000 000 094 994 902 518 784 × 2 = 0 + 0.000 000 189 989 805 037 568;
  • 9) 0.000 000 189 989 805 037 568 × 2 = 0 + 0.000 000 379 979 610 075 136;
  • 10) 0.000 000 379 979 610 075 136 × 2 = 0 + 0.000 000 759 959 220 150 272;
  • 11) 0.000 000 759 959 220 150 272 × 2 = 0 + 0.000 001 519 918 440 300 544;
  • 12) 0.000 001 519 918 440 300 544 × 2 = 0 + 0.000 003 039 836 880 601 088;
  • 13) 0.000 003 039 836 880 601 088 × 2 = 0 + 0.000 006 079 673 761 202 176;
  • 14) 0.000 006 079 673 761 202 176 × 2 = 0 + 0.000 012 159 347 522 404 352;
  • 15) 0.000 012 159 347 522 404 352 × 2 = 0 + 0.000 024 318 695 044 808 704;
  • 16) 0.000 024 318 695 044 808 704 × 2 = 0 + 0.000 048 637 390 089 617 408;
  • 17) 0.000 048 637 390 089 617 408 × 2 = 0 + 0.000 097 274 780 179 234 816;
  • 18) 0.000 097 274 780 179 234 816 × 2 = 0 + 0.000 194 549 560 358 469 632;
  • 19) 0.000 194 549 560 358 469 632 × 2 = 0 + 0.000 389 099 120 716 939 264;
  • 20) 0.000 389 099 120 716 939 264 × 2 = 0 + 0.000 778 198 241 433 878 528;
  • 21) 0.000 778 198 241 433 878 528 × 2 = 0 + 0.001 556 396 482 867 757 056;
  • 22) 0.001 556 396 482 867 757 056 × 2 = 0 + 0.003 112 792 965 735 514 112;
  • 23) 0.003 112 792 965 735 514 112 × 2 = 0 + 0.006 225 585 931 471 028 224;
  • 24) 0.006 225 585 931 471 028 224 × 2 = 0 + 0.012 451 171 862 942 056 448;
  • 25) 0.012 451 171 862 942 056 448 × 2 = 0 + 0.024 902 343 725 884 112 896;
  • 26) 0.024 902 343 725 884 112 896 × 2 = 0 + 0.049 804 687 451 768 225 792;
  • 27) 0.049 804 687 451 768 225 792 × 2 = 0 + 0.099 609 374 903 536 451 584;
  • 28) 0.099 609 374 903 536 451 584 × 2 = 0 + 0.199 218 749 807 072 903 168;
  • 29) 0.199 218 749 807 072 903 168 × 2 = 0 + 0.398 437 499 614 145 806 336;
  • 30) 0.398 437 499 614 145 806 336 × 2 = 0 + 0.796 874 999 228 291 612 672;
  • 31) 0.796 874 999 228 291 612 672 × 2 = 1 + 0.593 749 998 456 583 225 344;
  • 32) 0.593 749 998 456 583 225 344 × 2 = 1 + 0.187 499 996 913 166 450 688;
  • 33) 0.187 499 996 913 166 450 688 × 2 = 0 + 0.374 999 993 826 332 901 376;
  • 34) 0.374 999 993 826 332 901 376 × 2 = 0 + 0.749 999 987 652 665 802 752;
  • 35) 0.749 999 987 652 665 802 752 × 2 = 1 + 0.499 999 975 305 331 605 504;
  • 36) 0.499 999 975 305 331 605 504 × 2 = 0 + 0.999 999 950 610 663 211 008;
  • 37) 0.999 999 950 610 663 211 008 × 2 = 1 + 0.999 999 901 221 326 422 016;
  • 38) 0.999 999 901 221 326 422 016 × 2 = 1 + 0.999 999 802 442 652 844 032;
  • 39) 0.999 999 802 442 652 844 032 × 2 = 1 + 0.999 999 604 885 305 688 064;
  • 40) 0.999 999 604 885 305 688 064 × 2 = 1 + 0.999 999 209 770 611 376 128;
  • 41) 0.999 999 209 770 611 376 128 × 2 = 1 + 0.999 998 419 541 222 752 256;
  • 42) 0.999 998 419 541 222 752 256 × 2 = 1 + 0.999 996 839 082 445 504 512;
  • 43) 0.999 996 839 082 445 504 512 × 2 = 1 + 0.999 993 678 164 891 009 024;
  • 44) 0.999 993 678 164 891 009 024 × 2 = 1 + 0.999 987 356 329 782 018 048;
  • 45) 0.999 987 356 329 782 018 048 × 2 = 1 + 0.999 974 712 659 564 036 096;
  • 46) 0.999 974 712 659 564 036 096 × 2 = 1 + 0.999 949 425 319 128 072 192;
  • 47) 0.999 949 425 319 128 072 192 × 2 = 1 + 0.999 898 850 638 256 144 384;
  • 48) 0.999 898 850 638 256 144 384 × 2 = 1 + 0.999 797 701 276 512 288 768;
  • 49) 0.999 797 701 276 512 288 768 × 2 = 1 + 0.999 595 402 553 024 577 536;
  • 50) 0.999 595 402 553 024 577 536 × 2 = 1 + 0.999 190 805 106 049 155 072;
  • 51) 0.999 190 805 106 049 155 072 × 2 = 1 + 0.998 381 610 212 098 310 144;
  • 52) 0.998 381 610 212 098 310 144 × 2 = 1 + 0.996 763 220 424 196 620 288;
  • 53) 0.996 763 220 424 196 620 288 × 2 = 1 + 0.993 526 440 848 393 240 576;
  • 54) 0.993 526 440 848 393 240 576 × 2 = 1 + 0.987 052 881 696 786 481 152;

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:


0.000 000 000 742 147 675 928(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 928(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:


0.000 000 000 742 147 675 928(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 -0.000 000 000 742 147 675 928 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