-0.000 000 000 742 147 676 646 697 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 697 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
-0.000 000 000 742 147 676 646 697 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:

|-0.000 000 000 742 147 676 646 697 8| = 0.000 000 000 742 147 676 646 697 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: 0.000 000 000 742 147 676 646 697 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) 0.000 000 000 742 147 676 646 697 8 × 2 = 0 + 0.000 000 001 484 295 353 293 395 6;
  • 2) 0.000 000 001 484 295 353 293 395 6 × 2 = 0 + 0.000 000 002 968 590 706 586 791 2;
  • 3) 0.000 000 002 968 590 706 586 791 2 × 2 = 0 + 0.000 000 005 937 181 413 173 582 4;
  • 4) 0.000 000 005 937 181 413 173 582 4 × 2 = 0 + 0.000 000 011 874 362 826 347 164 8;
  • 5) 0.000 000 011 874 362 826 347 164 8 × 2 = 0 + 0.000 000 023 748 725 652 694 329 6;
  • 6) 0.000 000 023 748 725 652 694 329 6 × 2 = 0 + 0.000 000 047 497 451 305 388 659 2;
  • 7) 0.000 000 047 497 451 305 388 659 2 × 2 = 0 + 0.000 000 094 994 902 610 777 318 4;
  • 8) 0.000 000 094 994 902 610 777 318 4 × 2 = 0 + 0.000 000 189 989 805 221 554 636 8;
  • 9) 0.000 000 189 989 805 221 554 636 8 × 2 = 0 + 0.000 000 379 979 610 443 109 273 6;
  • 10) 0.000 000 379 979 610 443 109 273 6 × 2 = 0 + 0.000 000 759 959 220 886 218 547 2;
  • 11) 0.000 000 759 959 220 886 218 547 2 × 2 = 0 + 0.000 001 519 918 441 772 437 094 4;
  • 12) 0.000 001 519 918 441 772 437 094 4 × 2 = 0 + 0.000 003 039 836 883 544 874 188 8;
  • 13) 0.000 003 039 836 883 544 874 188 8 × 2 = 0 + 0.000 006 079 673 767 089 748 377 6;
  • 14) 0.000 006 079 673 767 089 748 377 6 × 2 = 0 + 0.000 012 159 347 534 179 496 755 2;
  • 15) 0.000 012 159 347 534 179 496 755 2 × 2 = 0 + 0.000 024 318 695 068 358 993 510 4;
  • 16) 0.000 024 318 695 068 358 993 510 4 × 2 = 0 + 0.000 048 637 390 136 717 987 020 8;
  • 17) 0.000 048 637 390 136 717 987 020 8 × 2 = 0 + 0.000 097 274 780 273 435 974 041 6;
  • 18) 0.000 097 274 780 273 435 974 041 6 × 2 = 0 + 0.000 194 549 560 546 871 948 083 2;
  • 19) 0.000 194 549 560 546 871 948 083 2 × 2 = 0 + 0.000 389 099 121 093 743 896 166 4;
  • 20) 0.000 389 099 121 093 743 896 166 4 × 2 = 0 + 0.000 778 198 242 187 487 792 332 8;
  • 21) 0.000 778 198 242 187 487 792 332 8 × 2 = 0 + 0.001 556 396 484 374 975 584 665 6;
  • 22) 0.001 556 396 484 374 975 584 665 6 × 2 = 0 + 0.003 112 792 968 749 951 169 331 2;
  • 23) 0.003 112 792 968 749 951 169 331 2 × 2 = 0 + 0.006 225 585 937 499 902 338 662 4;
  • 24) 0.006 225 585 937 499 902 338 662 4 × 2 = 0 + 0.012 451 171 874 999 804 677 324 8;
  • 25) 0.012 451 171 874 999 804 677 324 8 × 2 = 0 + 0.024 902 343 749 999 609 354 649 6;
  • 26) 0.024 902 343 749 999 609 354 649 6 × 2 = 0 + 0.049 804 687 499 999 218 709 299 2;
  • 27) 0.049 804 687 499 999 218 709 299 2 × 2 = 0 + 0.099 609 374 999 998 437 418 598 4;
  • 28) 0.099 609 374 999 998 437 418 598 4 × 2 = 0 + 0.199 218 749 999 996 874 837 196 8;
  • 29) 0.199 218 749 999 996 874 837 196 8 × 2 = 0 + 0.398 437 499 999 993 749 674 393 6;
  • 30) 0.398 437 499 999 993 749 674 393 6 × 2 = 0 + 0.796 874 999 999 987 499 348 787 2;
  • 31) 0.796 874 999 999 987 499 348 787 2 × 2 = 1 + 0.593 749 999 999 974 998 697 574 4;
  • 32) 0.593 749 999 999 974 998 697 574 4 × 2 = 1 + 0.187 499 999 999 949 997 395 148 8;
  • 33) 0.187 499 999 999 949 997 395 148 8 × 2 = 0 + 0.374 999 999 999 899 994 790 297 6;
  • 34) 0.374 999 999 999 899 994 790 297 6 × 2 = 0 + 0.749 999 999 999 799 989 580 595 2;
  • 35) 0.749 999 999 999 799 989 580 595 2 × 2 = 1 + 0.499 999 999 999 599 979 161 190 4;
  • 36) 0.499 999 999 999 599 979 161 190 4 × 2 = 0 + 0.999 999 999 999 199 958 322 380 8;
  • 37) 0.999 999 999 999 199 958 322 380 8 × 2 = 1 + 0.999 999 999 998 399 916 644 761 6;
  • 38) 0.999 999 999 998 399 916 644 761 6 × 2 = 1 + 0.999 999 999 996 799 833 289 523 2;
  • 39) 0.999 999 999 996 799 833 289 523 2 × 2 = 1 + 0.999 999 999 993 599 666 579 046 4;
  • 40) 0.999 999 999 993 599 666 579 046 4 × 2 = 1 + 0.999 999 999 987 199 333 158 092 8;
  • 41) 0.999 999 999 987 199 333 158 092 8 × 2 = 1 + 0.999 999 999 974 398 666 316 185 6;
  • 42) 0.999 999 999 974 398 666 316 185 6 × 2 = 1 + 0.999 999 999 948 797 332 632 371 2;
  • 43) 0.999 999 999 948 797 332 632 371 2 × 2 = 1 + 0.999 999 999 897 594 665 264 742 4;
  • 44) 0.999 999 999 897 594 665 264 742 4 × 2 = 1 + 0.999 999 999 795 189 330 529 484 8;
  • 45) 0.999 999 999 795 189 330 529 484 8 × 2 = 1 + 0.999 999 999 590 378 661 058 969 6;
  • 46) 0.999 999 999 590 378 661 058 969 6 × 2 = 1 + 0.999 999 999 180 757 322 117 939 2;
  • 47) 0.999 999 999 180 757 322 117 939 2 × 2 = 1 + 0.999 999 998 361 514 644 235 878 4;
  • 48) 0.999 999 998 361 514 644 235 878 4 × 2 = 1 + 0.999 999 996 723 029 288 471 756 8;
  • 49) 0.999 999 996 723 029 288 471 756 8 × 2 = 1 + 0.999 999 993 446 058 576 943 513 6;
  • 50) 0.999 999 993 446 058 576 943 513 6 × 2 = 1 + 0.999 999 986 892 117 153 887 027 2;
  • 51) 0.999 999 986 892 117 153 887 027 2 × 2 = 1 + 0.999 999 973 784 234 307 774 054 4;
  • 52) 0.999 999 973 784 234 307 774 054 4 × 2 = 1 + 0.999 999 947 568 468 615 548 108 8;
  • 53) 0.999 999 947 568 468 615 548 108 8 × 2 = 1 + 0.999 999 895 136 937 231 096 217 6;
  • 54) 0.999 999 895 136 937 231 096 217 6 × 2 = 1 + 0.999 999 790 273 874 462 192 435 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:


0.000 000 000 742 147 676 646 697 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 697 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:


0.000 000 000 742 147 676 646 697 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 -0.000 000 000 742 147 676 646 697 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 -0.000 000 000 742 147 676 646 692 3 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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