-0.000 000 000 742 147 676 289 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 289(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
-0.000 000 000 742 147 676 289(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:

|-0.000 000 000 742 147 676 289| = 0.000 000 000 742 147 676 289


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: 0.000 000 000 742 147 676 289.

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) 0.000 000 000 742 147 676 289 × 2 = 0 + 0.000 000 001 484 295 352 578;
  • 2) 0.000 000 001 484 295 352 578 × 2 = 0 + 0.000 000 002 968 590 705 156;
  • 3) 0.000 000 002 968 590 705 156 × 2 = 0 + 0.000 000 005 937 181 410 312;
  • 4) 0.000 000 005 937 181 410 312 × 2 = 0 + 0.000 000 011 874 362 820 624;
  • 5) 0.000 000 011 874 362 820 624 × 2 = 0 + 0.000 000 023 748 725 641 248;
  • 6) 0.000 000 023 748 725 641 248 × 2 = 0 + 0.000 000 047 497 451 282 496;
  • 7) 0.000 000 047 497 451 282 496 × 2 = 0 + 0.000 000 094 994 902 564 992;
  • 8) 0.000 000 094 994 902 564 992 × 2 = 0 + 0.000 000 189 989 805 129 984;
  • 9) 0.000 000 189 989 805 129 984 × 2 = 0 + 0.000 000 379 979 610 259 968;
  • 10) 0.000 000 379 979 610 259 968 × 2 = 0 + 0.000 000 759 959 220 519 936;
  • 11) 0.000 000 759 959 220 519 936 × 2 = 0 + 0.000 001 519 918 441 039 872;
  • 12) 0.000 001 519 918 441 039 872 × 2 = 0 + 0.000 003 039 836 882 079 744;
  • 13) 0.000 003 039 836 882 079 744 × 2 = 0 + 0.000 006 079 673 764 159 488;
  • 14) 0.000 006 079 673 764 159 488 × 2 = 0 + 0.000 012 159 347 528 318 976;
  • 15) 0.000 012 159 347 528 318 976 × 2 = 0 + 0.000 024 318 695 056 637 952;
  • 16) 0.000 024 318 695 056 637 952 × 2 = 0 + 0.000 048 637 390 113 275 904;
  • 17) 0.000 048 637 390 113 275 904 × 2 = 0 + 0.000 097 274 780 226 551 808;
  • 18) 0.000 097 274 780 226 551 808 × 2 = 0 + 0.000 194 549 560 453 103 616;
  • 19) 0.000 194 549 560 453 103 616 × 2 = 0 + 0.000 389 099 120 906 207 232;
  • 20) 0.000 389 099 120 906 207 232 × 2 = 0 + 0.000 778 198 241 812 414 464;
  • 21) 0.000 778 198 241 812 414 464 × 2 = 0 + 0.001 556 396 483 624 828 928;
  • 22) 0.001 556 396 483 624 828 928 × 2 = 0 + 0.003 112 792 967 249 657 856;
  • 23) 0.003 112 792 967 249 657 856 × 2 = 0 + 0.006 225 585 934 499 315 712;
  • 24) 0.006 225 585 934 499 315 712 × 2 = 0 + 0.012 451 171 868 998 631 424;
  • 25) 0.012 451 171 868 998 631 424 × 2 = 0 + 0.024 902 343 737 997 262 848;
  • 26) 0.024 902 343 737 997 262 848 × 2 = 0 + 0.049 804 687 475 994 525 696;
  • 27) 0.049 804 687 475 994 525 696 × 2 = 0 + 0.099 609 374 951 989 051 392;
  • 28) 0.099 609 374 951 989 051 392 × 2 = 0 + 0.199 218 749 903 978 102 784;
  • 29) 0.199 218 749 903 978 102 784 × 2 = 0 + 0.398 437 499 807 956 205 568;
  • 30) 0.398 437 499 807 956 205 568 × 2 = 0 + 0.796 874 999 615 912 411 136;
  • 31) 0.796 874 999 615 912 411 136 × 2 = 1 + 0.593 749 999 231 824 822 272;
  • 32) 0.593 749 999 231 824 822 272 × 2 = 1 + 0.187 499 998 463 649 644 544;
  • 33) 0.187 499 998 463 649 644 544 × 2 = 0 + 0.374 999 996 927 299 289 088;
  • 34) 0.374 999 996 927 299 289 088 × 2 = 0 + 0.749 999 993 854 598 578 176;
  • 35) 0.749 999 993 854 598 578 176 × 2 = 1 + 0.499 999 987 709 197 156 352;
  • 36) 0.499 999 987 709 197 156 352 × 2 = 0 + 0.999 999 975 418 394 312 704;
  • 37) 0.999 999 975 418 394 312 704 × 2 = 1 + 0.999 999 950 836 788 625 408;
  • 38) 0.999 999 950 836 788 625 408 × 2 = 1 + 0.999 999 901 673 577 250 816;
  • 39) 0.999 999 901 673 577 250 816 × 2 = 1 + 0.999 999 803 347 154 501 632;
  • 40) 0.999 999 803 347 154 501 632 × 2 = 1 + 0.999 999 606 694 309 003 264;
  • 41) 0.999 999 606 694 309 003 264 × 2 = 1 + 0.999 999 213 388 618 006 528;
  • 42) 0.999 999 213 388 618 006 528 × 2 = 1 + 0.999 998 426 777 236 013 056;
  • 43) 0.999 998 426 777 236 013 056 × 2 = 1 + 0.999 996 853 554 472 026 112;
  • 44) 0.999 996 853 554 472 026 112 × 2 = 1 + 0.999 993 707 108 944 052 224;
  • 45) 0.999 993 707 108 944 052 224 × 2 = 1 + 0.999 987 414 217 888 104 448;
  • 46) 0.999 987 414 217 888 104 448 × 2 = 1 + 0.999 974 828 435 776 208 896;
  • 47) 0.999 974 828 435 776 208 896 × 2 = 1 + 0.999 949 656 871 552 417 792;
  • 48) 0.999 949 656 871 552 417 792 × 2 = 1 + 0.999 899 313 743 104 835 584;
  • 49) 0.999 899 313 743 104 835 584 × 2 = 1 + 0.999 798 627 486 209 671 168;
  • 50) 0.999 798 627 486 209 671 168 × 2 = 1 + 0.999 597 254 972 419 342 336;
  • 51) 0.999 597 254 972 419 342 336 × 2 = 1 + 0.999 194 509 944 838 684 672;
  • 52) 0.999 194 509 944 838 684 672 × 2 = 1 + 0.998 389 019 889 677 369 344;
  • 53) 0.998 389 019 889 677 369 344 × 2 = 1 + 0.996 778 039 779 354 738 688;
  • 54) 0.996 778 039 779 354 738 688 × 2 = 1 + 0.993 556 079 558 709 477 376;

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:


0.000 000 000 742 147 676 289(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 289(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:


0.000 000 000 742 147 676 289(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 -0.000 000 000 742 147 676 289 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 233 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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