-0.000 000 000 742 147 689 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 689 8(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
-0.000 000 000 742 147 689 8(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:

|-0.000 000 000 742 147 689 8| = 0.000 000 000 742 147 689 8


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: 0.000 000 000 742 147 689 8.

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) 0.000 000 000 742 147 689 8 × 2 = 0 + 0.000 000 001 484 295 379 6;
  • 2) 0.000 000 001 484 295 379 6 × 2 = 0 + 0.000 000 002 968 590 759 2;
  • 3) 0.000 000 002 968 590 759 2 × 2 = 0 + 0.000 000 005 937 181 518 4;
  • 4) 0.000 000 005 937 181 518 4 × 2 = 0 + 0.000 000 011 874 363 036 8;
  • 5) 0.000 000 011 874 363 036 8 × 2 = 0 + 0.000 000 023 748 726 073 6;
  • 6) 0.000 000 023 748 726 073 6 × 2 = 0 + 0.000 000 047 497 452 147 2;
  • 7) 0.000 000 047 497 452 147 2 × 2 = 0 + 0.000 000 094 994 904 294 4;
  • 8) 0.000 000 094 994 904 294 4 × 2 = 0 + 0.000 000 189 989 808 588 8;
  • 9) 0.000 000 189 989 808 588 8 × 2 = 0 + 0.000 000 379 979 617 177 6;
  • 10) 0.000 000 379 979 617 177 6 × 2 = 0 + 0.000 000 759 959 234 355 2;
  • 11) 0.000 000 759 959 234 355 2 × 2 = 0 + 0.000 001 519 918 468 710 4;
  • 12) 0.000 001 519 918 468 710 4 × 2 = 0 + 0.000 003 039 836 937 420 8;
  • 13) 0.000 003 039 836 937 420 8 × 2 = 0 + 0.000 006 079 673 874 841 6;
  • 14) 0.000 006 079 673 874 841 6 × 2 = 0 + 0.000 012 159 347 749 683 2;
  • 15) 0.000 012 159 347 749 683 2 × 2 = 0 + 0.000 024 318 695 499 366 4;
  • 16) 0.000 024 318 695 499 366 4 × 2 = 0 + 0.000 048 637 390 998 732 8;
  • 17) 0.000 048 637 390 998 732 8 × 2 = 0 + 0.000 097 274 781 997 465 6;
  • 18) 0.000 097 274 781 997 465 6 × 2 = 0 + 0.000 194 549 563 994 931 2;
  • 19) 0.000 194 549 563 994 931 2 × 2 = 0 + 0.000 389 099 127 989 862 4;
  • 20) 0.000 389 099 127 989 862 4 × 2 = 0 + 0.000 778 198 255 979 724 8;
  • 21) 0.000 778 198 255 979 724 8 × 2 = 0 + 0.001 556 396 511 959 449 6;
  • 22) 0.001 556 396 511 959 449 6 × 2 = 0 + 0.003 112 793 023 918 899 2;
  • 23) 0.003 112 793 023 918 899 2 × 2 = 0 + 0.006 225 586 047 837 798 4;
  • 24) 0.006 225 586 047 837 798 4 × 2 = 0 + 0.012 451 172 095 675 596 8;
  • 25) 0.012 451 172 095 675 596 8 × 2 = 0 + 0.024 902 344 191 351 193 6;
  • 26) 0.024 902 344 191 351 193 6 × 2 = 0 + 0.049 804 688 382 702 387 2;
  • 27) 0.049 804 688 382 702 387 2 × 2 = 0 + 0.099 609 376 765 404 774 4;
  • 28) 0.099 609 376 765 404 774 4 × 2 = 0 + 0.199 218 753 530 809 548 8;
  • 29) 0.199 218 753 530 809 548 8 × 2 = 0 + 0.398 437 507 061 619 097 6;
  • 30) 0.398 437 507 061 619 097 6 × 2 = 0 + 0.796 875 014 123 238 195 2;
  • 31) 0.796 875 014 123 238 195 2 × 2 = 1 + 0.593 750 028 246 476 390 4;
  • 32) 0.593 750 028 246 476 390 4 × 2 = 1 + 0.187 500 056 492 952 780 8;
  • 33) 0.187 500 056 492 952 780 8 × 2 = 0 + 0.375 000 112 985 905 561 6;
  • 34) 0.375 000 112 985 905 561 6 × 2 = 0 + 0.750 000 225 971 811 123 2;
  • 35) 0.750 000 225 971 811 123 2 × 2 = 1 + 0.500 000 451 943 622 246 4;
  • 36) 0.500 000 451 943 622 246 4 × 2 = 1 + 0.000 000 903 887 244 492 8;
  • 37) 0.000 000 903 887 244 492 8 × 2 = 0 + 0.000 001 807 774 488 985 6;
  • 38) 0.000 001 807 774 488 985 6 × 2 = 0 + 0.000 003 615 548 977 971 2;
  • 39) 0.000 003 615 548 977 971 2 × 2 = 0 + 0.000 007 231 097 955 942 4;
  • 40) 0.000 007 231 097 955 942 4 × 2 = 0 + 0.000 014 462 195 911 884 8;
  • 41) 0.000 014 462 195 911 884 8 × 2 = 0 + 0.000 028 924 391 823 769 6;
  • 42) 0.000 028 924 391 823 769 6 × 2 = 0 + 0.000 057 848 783 647 539 2;
  • 43) 0.000 057 848 783 647 539 2 × 2 = 0 + 0.000 115 697 567 295 078 4;
  • 44) 0.000 115 697 567 295 078 4 × 2 = 0 + 0.000 231 395 134 590 156 8;
  • 45) 0.000 231 395 134 590 156 8 × 2 = 0 + 0.000 462 790 269 180 313 6;
  • 46) 0.000 462 790 269 180 313 6 × 2 = 0 + 0.000 925 580 538 360 627 2;
  • 47) 0.000 925 580 538 360 627 2 × 2 = 0 + 0.001 851 161 076 721 254 4;
  • 48) 0.001 851 161 076 721 254 4 × 2 = 0 + 0.003 702 322 153 442 508 8;
  • 49) 0.003 702 322 153 442 508 8 × 2 = 0 + 0.007 404 644 306 885 017 6;
  • 50) 0.007 404 644 306 885 017 6 × 2 = 0 + 0.014 809 288 613 770 035 2;
  • 51) 0.014 809 288 613 770 035 2 × 2 = 0 + 0.029 618 577 227 540 070 4;
  • 52) 0.029 618 577 227 540 070 4 × 2 = 0 + 0.059 237 154 455 080 140 8;
  • 53) 0.059 237 154 455 080 140 8 × 2 = 0 + 0.118 474 308 910 160 281 6;
  • 54) 0.118 474 308 910 160 281 6 × 2 = 0 + 0.236 948 617 820 320 563 2;

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:


0.000 000 000 742 147 689 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 689 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 689 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 689 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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