-0.000 000 000 742 147 679 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 679 6(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
-0.000 000 000 742 147 679 6(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:

|-0.000 000 000 742 147 679 6| = 0.000 000 000 742 147 679 6


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: 0.000 000 000 742 147 679 6.

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) 0.000 000 000 742 147 679 6 × 2 = 0 + 0.000 000 001 484 295 359 2;
  • 2) 0.000 000 001 484 295 359 2 × 2 = 0 + 0.000 000 002 968 590 718 4;
  • 3) 0.000 000 002 968 590 718 4 × 2 = 0 + 0.000 000 005 937 181 436 8;
  • 4) 0.000 000 005 937 181 436 8 × 2 = 0 + 0.000 000 011 874 362 873 6;
  • 5) 0.000 000 011 874 362 873 6 × 2 = 0 + 0.000 000 023 748 725 747 2;
  • 6) 0.000 000 023 748 725 747 2 × 2 = 0 + 0.000 000 047 497 451 494 4;
  • 7) 0.000 000 047 497 451 494 4 × 2 = 0 + 0.000 000 094 994 902 988 8;
  • 8) 0.000 000 094 994 902 988 8 × 2 = 0 + 0.000 000 189 989 805 977 6;
  • 9) 0.000 000 189 989 805 977 6 × 2 = 0 + 0.000 000 379 979 611 955 2;
  • 10) 0.000 000 379 979 611 955 2 × 2 = 0 + 0.000 000 759 959 223 910 4;
  • 11) 0.000 000 759 959 223 910 4 × 2 = 0 + 0.000 001 519 918 447 820 8;
  • 12) 0.000 001 519 918 447 820 8 × 2 = 0 + 0.000 003 039 836 895 641 6;
  • 13) 0.000 003 039 836 895 641 6 × 2 = 0 + 0.000 006 079 673 791 283 2;
  • 14) 0.000 006 079 673 791 283 2 × 2 = 0 + 0.000 012 159 347 582 566 4;
  • 15) 0.000 012 159 347 582 566 4 × 2 = 0 + 0.000 024 318 695 165 132 8;
  • 16) 0.000 024 318 695 165 132 8 × 2 = 0 + 0.000 048 637 390 330 265 6;
  • 17) 0.000 048 637 390 330 265 6 × 2 = 0 + 0.000 097 274 780 660 531 2;
  • 18) 0.000 097 274 780 660 531 2 × 2 = 0 + 0.000 194 549 561 321 062 4;
  • 19) 0.000 194 549 561 321 062 4 × 2 = 0 + 0.000 389 099 122 642 124 8;
  • 20) 0.000 389 099 122 642 124 8 × 2 = 0 + 0.000 778 198 245 284 249 6;
  • 21) 0.000 778 198 245 284 249 6 × 2 = 0 + 0.001 556 396 490 568 499 2;
  • 22) 0.001 556 396 490 568 499 2 × 2 = 0 + 0.003 112 792 981 136 998 4;
  • 23) 0.003 112 792 981 136 998 4 × 2 = 0 + 0.006 225 585 962 273 996 8;
  • 24) 0.006 225 585 962 273 996 8 × 2 = 0 + 0.012 451 171 924 547 993 6;
  • 25) 0.012 451 171 924 547 993 6 × 2 = 0 + 0.024 902 343 849 095 987 2;
  • 26) 0.024 902 343 849 095 987 2 × 2 = 0 + 0.049 804 687 698 191 974 4;
  • 27) 0.049 804 687 698 191 974 4 × 2 = 0 + 0.099 609 375 396 383 948 8;
  • 28) 0.099 609 375 396 383 948 8 × 2 = 0 + 0.199 218 750 792 767 897 6;
  • 29) 0.199 218 750 792 767 897 6 × 2 = 0 + 0.398 437 501 585 535 795 2;
  • 30) 0.398 437 501 585 535 795 2 × 2 = 0 + 0.796 875 003 171 071 590 4;
  • 31) 0.796 875 003 171 071 590 4 × 2 = 1 + 0.593 750 006 342 143 180 8;
  • 32) 0.593 750 006 342 143 180 8 × 2 = 1 + 0.187 500 012 684 286 361 6;
  • 33) 0.187 500 012 684 286 361 6 × 2 = 0 + 0.375 000 025 368 572 723 2;
  • 34) 0.375 000 025 368 572 723 2 × 2 = 0 + 0.750 000 050 737 145 446 4;
  • 35) 0.750 000 050 737 145 446 4 × 2 = 1 + 0.500 000 101 474 290 892 8;
  • 36) 0.500 000 101 474 290 892 8 × 2 = 1 + 0.000 000 202 948 581 785 6;
  • 37) 0.000 000 202 948 581 785 6 × 2 = 0 + 0.000 000 405 897 163 571 2;
  • 38) 0.000 000 405 897 163 571 2 × 2 = 0 + 0.000 000 811 794 327 142 4;
  • 39) 0.000 000 811 794 327 142 4 × 2 = 0 + 0.000 001 623 588 654 284 8;
  • 40) 0.000 001 623 588 654 284 8 × 2 = 0 + 0.000 003 247 177 308 569 6;
  • 41) 0.000 003 247 177 308 569 6 × 2 = 0 + 0.000 006 494 354 617 139 2;
  • 42) 0.000 006 494 354 617 139 2 × 2 = 0 + 0.000 012 988 709 234 278 4;
  • 43) 0.000 012 988 709 234 278 4 × 2 = 0 + 0.000 025 977 418 468 556 8;
  • 44) 0.000 025 977 418 468 556 8 × 2 = 0 + 0.000 051 954 836 937 113 6;
  • 45) 0.000 051 954 836 937 113 6 × 2 = 0 + 0.000 103 909 673 874 227 2;
  • 46) 0.000 103 909 673 874 227 2 × 2 = 0 + 0.000 207 819 347 748 454 4;
  • 47) 0.000 207 819 347 748 454 4 × 2 = 0 + 0.000 415 638 695 496 908 8;
  • 48) 0.000 415 638 695 496 908 8 × 2 = 0 + 0.000 831 277 390 993 817 6;
  • 49) 0.000 831 277 390 993 817 6 × 2 = 0 + 0.001 662 554 781 987 635 2;
  • 50) 0.001 662 554 781 987 635 2 × 2 = 0 + 0.003 325 109 563 975 270 4;
  • 51) 0.003 325 109 563 975 270 4 × 2 = 0 + 0.006 650 219 127 950 540 8;
  • 52) 0.006 650 219 127 950 540 8 × 2 = 0 + 0.013 300 438 255 901 081 6;
  • 53) 0.013 300 438 255 901 081 6 × 2 = 0 + 0.026 600 876 511 802 163 2;
  • 54) 0.026 600 876 511 802 163 2 × 2 = 0 + 0.053 201 753 023 604 326 4;

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:


0.000 000 000 742 147 679 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 679 6(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:


0.000 000 000 742 147 679 6(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 -0.000 000 000 742 147 679 6 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