-0.000 000 000 742 147 664 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 664 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
-0.000 000 000 742 147 664 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:

|-0.000 000 000 742 147 664 8| = 0.000 000 000 742 147 664 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: 0.000 000 000 742 147 664 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) 0.000 000 000 742 147 664 8 × 2 = 0 + 0.000 000 001 484 295 329 6;
  • 2) 0.000 000 001 484 295 329 6 × 2 = 0 + 0.000 000 002 968 590 659 2;
  • 3) 0.000 000 002 968 590 659 2 × 2 = 0 + 0.000 000 005 937 181 318 4;
  • 4) 0.000 000 005 937 181 318 4 × 2 = 0 + 0.000 000 011 874 362 636 8;
  • 5) 0.000 000 011 874 362 636 8 × 2 = 0 + 0.000 000 023 748 725 273 6;
  • 6) 0.000 000 023 748 725 273 6 × 2 = 0 + 0.000 000 047 497 450 547 2;
  • 7) 0.000 000 047 497 450 547 2 × 2 = 0 + 0.000 000 094 994 901 094 4;
  • 8) 0.000 000 094 994 901 094 4 × 2 = 0 + 0.000 000 189 989 802 188 8;
  • 9) 0.000 000 189 989 802 188 8 × 2 = 0 + 0.000 000 379 979 604 377 6;
  • 10) 0.000 000 379 979 604 377 6 × 2 = 0 + 0.000 000 759 959 208 755 2;
  • 11) 0.000 000 759 959 208 755 2 × 2 = 0 + 0.000 001 519 918 417 510 4;
  • 12) 0.000 001 519 918 417 510 4 × 2 = 0 + 0.000 003 039 836 835 020 8;
  • 13) 0.000 003 039 836 835 020 8 × 2 = 0 + 0.000 006 079 673 670 041 6;
  • 14) 0.000 006 079 673 670 041 6 × 2 = 0 + 0.000 012 159 347 340 083 2;
  • 15) 0.000 012 159 347 340 083 2 × 2 = 0 + 0.000 024 318 694 680 166 4;
  • 16) 0.000 024 318 694 680 166 4 × 2 = 0 + 0.000 048 637 389 360 332 8;
  • 17) 0.000 048 637 389 360 332 8 × 2 = 0 + 0.000 097 274 778 720 665 6;
  • 18) 0.000 097 274 778 720 665 6 × 2 = 0 + 0.000 194 549 557 441 331 2;
  • 19) 0.000 194 549 557 441 331 2 × 2 = 0 + 0.000 389 099 114 882 662 4;
  • 20) 0.000 389 099 114 882 662 4 × 2 = 0 + 0.000 778 198 229 765 324 8;
  • 21) 0.000 778 198 229 765 324 8 × 2 = 0 + 0.001 556 396 459 530 649 6;
  • 22) 0.001 556 396 459 530 649 6 × 2 = 0 + 0.003 112 792 919 061 299 2;
  • 23) 0.003 112 792 919 061 299 2 × 2 = 0 + 0.006 225 585 838 122 598 4;
  • 24) 0.006 225 585 838 122 598 4 × 2 = 0 + 0.012 451 171 676 245 196 8;
  • 25) 0.012 451 171 676 245 196 8 × 2 = 0 + 0.024 902 343 352 490 393 6;
  • 26) 0.024 902 343 352 490 393 6 × 2 = 0 + 0.049 804 686 704 980 787 2;
  • 27) 0.049 804 686 704 980 787 2 × 2 = 0 + 0.099 609 373 409 961 574 4;
  • 28) 0.099 609 373 409 961 574 4 × 2 = 0 + 0.199 218 746 819 923 148 8;
  • 29) 0.199 218 746 819 923 148 8 × 2 = 0 + 0.398 437 493 639 846 297 6;
  • 30) 0.398 437 493 639 846 297 6 × 2 = 0 + 0.796 874 987 279 692 595 2;
  • 31) 0.796 874 987 279 692 595 2 × 2 = 1 + 0.593 749 974 559 385 190 4;
  • 32) 0.593 749 974 559 385 190 4 × 2 = 1 + 0.187 499 949 118 770 380 8;
  • 33) 0.187 499 949 118 770 380 8 × 2 = 0 + 0.374 999 898 237 540 761 6;
  • 34) 0.374 999 898 237 540 761 6 × 2 = 0 + 0.749 999 796 475 081 523 2;
  • 35) 0.749 999 796 475 081 523 2 × 2 = 1 + 0.499 999 592 950 163 046 4;
  • 36) 0.499 999 592 950 163 046 4 × 2 = 0 + 0.999 999 185 900 326 092 8;
  • 37) 0.999 999 185 900 326 092 8 × 2 = 1 + 0.999 998 371 800 652 185 6;
  • 38) 0.999 998 371 800 652 185 6 × 2 = 1 + 0.999 996 743 601 304 371 2;
  • 39) 0.999 996 743 601 304 371 2 × 2 = 1 + 0.999 993 487 202 608 742 4;
  • 40) 0.999 993 487 202 608 742 4 × 2 = 1 + 0.999 986 974 405 217 484 8;
  • 41) 0.999 986 974 405 217 484 8 × 2 = 1 + 0.999 973 948 810 434 969 6;
  • 42) 0.999 973 948 810 434 969 6 × 2 = 1 + 0.999 947 897 620 869 939 2;
  • 43) 0.999 947 897 620 869 939 2 × 2 = 1 + 0.999 895 795 241 739 878 4;
  • 44) 0.999 895 795 241 739 878 4 × 2 = 1 + 0.999 791 590 483 479 756 8;
  • 45) 0.999 791 590 483 479 756 8 × 2 = 1 + 0.999 583 180 966 959 513 6;
  • 46) 0.999 583 180 966 959 513 6 × 2 = 1 + 0.999 166 361 933 919 027 2;
  • 47) 0.999 166 361 933 919 027 2 × 2 = 1 + 0.998 332 723 867 838 054 4;
  • 48) 0.998 332 723 867 838 054 4 × 2 = 1 + 0.996 665 447 735 676 108 8;
  • 49) 0.996 665 447 735 676 108 8 × 2 = 1 + 0.993 330 895 471 352 217 6;
  • 50) 0.993 330 895 471 352 217 6 × 2 = 1 + 0.986 661 790 942 704 435 2;
  • 51) 0.986 661 790 942 704 435 2 × 2 = 1 + 0.973 323 581 885 408 870 4;
  • 52) 0.973 323 581 885 408 870 4 × 2 = 1 + 0.946 647 163 770 817 740 8;
  • 53) 0.946 647 163 770 817 740 8 × 2 = 1 + 0.893 294 327 541 635 481 6;
  • 54) 0.893 294 327 541 635 481 6 × 2 = 1 + 0.786 588 655 083 270 963 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:


0.000 000 000 742 147 664 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 664 8(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:


0.000 000 000 742 147 664 8(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 -0.000 000 000 742 147 664 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 666 5 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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