-0.000 000 000 742 147 676 635 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 635 4(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
-0.000 000 000 742 147 676 635 4(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:

|-0.000 000 000 742 147 676 635 4| = 0.000 000 000 742 147 676 635 4


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: 0.000 000 000 742 147 676 635 4.

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) 0.000 000 000 742 147 676 635 4 × 2 = 0 + 0.000 000 001 484 295 353 270 8;
  • 2) 0.000 000 001 484 295 353 270 8 × 2 = 0 + 0.000 000 002 968 590 706 541 6;
  • 3) 0.000 000 002 968 590 706 541 6 × 2 = 0 + 0.000 000 005 937 181 413 083 2;
  • 4) 0.000 000 005 937 181 413 083 2 × 2 = 0 + 0.000 000 011 874 362 826 166 4;
  • 5) 0.000 000 011 874 362 826 166 4 × 2 = 0 + 0.000 000 023 748 725 652 332 8;
  • 6) 0.000 000 023 748 725 652 332 8 × 2 = 0 + 0.000 000 047 497 451 304 665 6;
  • 7) 0.000 000 047 497 451 304 665 6 × 2 = 0 + 0.000 000 094 994 902 609 331 2;
  • 8) 0.000 000 094 994 902 609 331 2 × 2 = 0 + 0.000 000 189 989 805 218 662 4;
  • 9) 0.000 000 189 989 805 218 662 4 × 2 = 0 + 0.000 000 379 979 610 437 324 8;
  • 10) 0.000 000 379 979 610 437 324 8 × 2 = 0 + 0.000 000 759 959 220 874 649 6;
  • 11) 0.000 000 759 959 220 874 649 6 × 2 = 0 + 0.000 001 519 918 441 749 299 2;
  • 12) 0.000 001 519 918 441 749 299 2 × 2 = 0 + 0.000 003 039 836 883 498 598 4;
  • 13) 0.000 003 039 836 883 498 598 4 × 2 = 0 + 0.000 006 079 673 766 997 196 8;
  • 14) 0.000 006 079 673 766 997 196 8 × 2 = 0 + 0.000 012 159 347 533 994 393 6;
  • 15) 0.000 012 159 347 533 994 393 6 × 2 = 0 + 0.000 024 318 695 067 988 787 2;
  • 16) 0.000 024 318 695 067 988 787 2 × 2 = 0 + 0.000 048 637 390 135 977 574 4;
  • 17) 0.000 048 637 390 135 977 574 4 × 2 = 0 + 0.000 097 274 780 271 955 148 8;
  • 18) 0.000 097 274 780 271 955 148 8 × 2 = 0 + 0.000 194 549 560 543 910 297 6;
  • 19) 0.000 194 549 560 543 910 297 6 × 2 = 0 + 0.000 389 099 121 087 820 595 2;
  • 20) 0.000 389 099 121 087 820 595 2 × 2 = 0 + 0.000 778 198 242 175 641 190 4;
  • 21) 0.000 778 198 242 175 641 190 4 × 2 = 0 + 0.001 556 396 484 351 282 380 8;
  • 22) 0.001 556 396 484 351 282 380 8 × 2 = 0 + 0.003 112 792 968 702 564 761 6;
  • 23) 0.003 112 792 968 702 564 761 6 × 2 = 0 + 0.006 225 585 937 405 129 523 2;
  • 24) 0.006 225 585 937 405 129 523 2 × 2 = 0 + 0.012 451 171 874 810 259 046 4;
  • 25) 0.012 451 171 874 810 259 046 4 × 2 = 0 + 0.024 902 343 749 620 518 092 8;
  • 26) 0.024 902 343 749 620 518 092 8 × 2 = 0 + 0.049 804 687 499 241 036 185 6;
  • 27) 0.049 804 687 499 241 036 185 6 × 2 = 0 + 0.099 609 374 998 482 072 371 2;
  • 28) 0.099 609 374 998 482 072 371 2 × 2 = 0 + 0.199 218 749 996 964 144 742 4;
  • 29) 0.199 218 749 996 964 144 742 4 × 2 = 0 + 0.398 437 499 993 928 289 484 8;
  • 30) 0.398 437 499 993 928 289 484 8 × 2 = 0 + 0.796 874 999 987 856 578 969 6;
  • 31) 0.796 874 999 987 856 578 969 6 × 2 = 1 + 0.593 749 999 975 713 157 939 2;
  • 32) 0.593 749 999 975 713 157 939 2 × 2 = 1 + 0.187 499 999 951 426 315 878 4;
  • 33) 0.187 499 999 951 426 315 878 4 × 2 = 0 + 0.374 999 999 902 852 631 756 8;
  • 34) 0.374 999 999 902 852 631 756 8 × 2 = 0 + 0.749 999 999 805 705 263 513 6;
  • 35) 0.749 999 999 805 705 263 513 6 × 2 = 1 + 0.499 999 999 611 410 527 027 2;
  • 36) 0.499 999 999 611 410 527 027 2 × 2 = 0 + 0.999 999 999 222 821 054 054 4;
  • 37) 0.999 999 999 222 821 054 054 4 × 2 = 1 + 0.999 999 998 445 642 108 108 8;
  • 38) 0.999 999 998 445 642 108 108 8 × 2 = 1 + 0.999 999 996 891 284 216 217 6;
  • 39) 0.999 999 996 891 284 216 217 6 × 2 = 1 + 0.999 999 993 782 568 432 435 2;
  • 40) 0.999 999 993 782 568 432 435 2 × 2 = 1 + 0.999 999 987 565 136 864 870 4;
  • 41) 0.999 999 987 565 136 864 870 4 × 2 = 1 + 0.999 999 975 130 273 729 740 8;
  • 42) 0.999 999 975 130 273 729 740 8 × 2 = 1 + 0.999 999 950 260 547 459 481 6;
  • 43) 0.999 999 950 260 547 459 481 6 × 2 = 1 + 0.999 999 900 521 094 918 963 2;
  • 44) 0.999 999 900 521 094 918 963 2 × 2 = 1 + 0.999 999 801 042 189 837 926 4;
  • 45) 0.999 999 801 042 189 837 926 4 × 2 = 1 + 0.999 999 602 084 379 675 852 8;
  • 46) 0.999 999 602 084 379 675 852 8 × 2 = 1 + 0.999 999 204 168 759 351 705 6;
  • 47) 0.999 999 204 168 759 351 705 6 × 2 = 1 + 0.999 998 408 337 518 703 411 2;
  • 48) 0.999 998 408 337 518 703 411 2 × 2 = 1 + 0.999 996 816 675 037 406 822 4;
  • 49) 0.999 996 816 675 037 406 822 4 × 2 = 1 + 0.999 993 633 350 074 813 644 8;
  • 50) 0.999 993 633 350 074 813 644 8 × 2 = 1 + 0.999 987 266 700 149 627 289 6;
  • 51) 0.999 987 266 700 149 627 289 6 × 2 = 1 + 0.999 974 533 400 299 254 579 2;
  • 52) 0.999 974 533 400 299 254 579 2 × 2 = 1 + 0.999 949 066 800 598 509 158 4;
  • 53) 0.999 949 066 800 598 509 158 4 × 2 = 1 + 0.999 898 133 601 197 018 316 8;
  • 54) 0.999 898 133 601 197 018 316 8 × 2 = 1 + 0.999 796 267 202 394 036 633 6;

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:


0.000 000 000 742 147 676 635 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 635 4(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:


0.000 000 000 742 147 676 635 4(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 -0.000 000 000 742 147 676 635 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 639 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: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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