-0.000 000 000 742 147 676 323 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 323(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
-0.000 000 000 742 147 676 323(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:

|-0.000 000 000 742 147 676 323| = 0.000 000 000 742 147 676 323


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: 0.000 000 000 742 147 676 323.

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) 0.000 000 000 742 147 676 323 × 2 = 0 + 0.000 000 001 484 295 352 646;
  • 2) 0.000 000 001 484 295 352 646 × 2 = 0 + 0.000 000 002 968 590 705 292;
  • 3) 0.000 000 002 968 590 705 292 × 2 = 0 + 0.000 000 005 937 181 410 584;
  • 4) 0.000 000 005 937 181 410 584 × 2 = 0 + 0.000 000 011 874 362 821 168;
  • 5) 0.000 000 011 874 362 821 168 × 2 = 0 + 0.000 000 023 748 725 642 336;
  • 6) 0.000 000 023 748 725 642 336 × 2 = 0 + 0.000 000 047 497 451 284 672;
  • 7) 0.000 000 047 497 451 284 672 × 2 = 0 + 0.000 000 094 994 902 569 344;
  • 8) 0.000 000 094 994 902 569 344 × 2 = 0 + 0.000 000 189 989 805 138 688;
  • 9) 0.000 000 189 989 805 138 688 × 2 = 0 + 0.000 000 379 979 610 277 376;
  • 10) 0.000 000 379 979 610 277 376 × 2 = 0 + 0.000 000 759 959 220 554 752;
  • 11) 0.000 000 759 959 220 554 752 × 2 = 0 + 0.000 001 519 918 441 109 504;
  • 12) 0.000 001 519 918 441 109 504 × 2 = 0 + 0.000 003 039 836 882 219 008;
  • 13) 0.000 003 039 836 882 219 008 × 2 = 0 + 0.000 006 079 673 764 438 016;
  • 14) 0.000 006 079 673 764 438 016 × 2 = 0 + 0.000 012 159 347 528 876 032;
  • 15) 0.000 012 159 347 528 876 032 × 2 = 0 + 0.000 024 318 695 057 752 064;
  • 16) 0.000 024 318 695 057 752 064 × 2 = 0 + 0.000 048 637 390 115 504 128;
  • 17) 0.000 048 637 390 115 504 128 × 2 = 0 + 0.000 097 274 780 231 008 256;
  • 18) 0.000 097 274 780 231 008 256 × 2 = 0 + 0.000 194 549 560 462 016 512;
  • 19) 0.000 194 549 560 462 016 512 × 2 = 0 + 0.000 389 099 120 924 033 024;
  • 20) 0.000 389 099 120 924 033 024 × 2 = 0 + 0.000 778 198 241 848 066 048;
  • 21) 0.000 778 198 241 848 066 048 × 2 = 0 + 0.001 556 396 483 696 132 096;
  • 22) 0.001 556 396 483 696 132 096 × 2 = 0 + 0.003 112 792 967 392 264 192;
  • 23) 0.003 112 792 967 392 264 192 × 2 = 0 + 0.006 225 585 934 784 528 384;
  • 24) 0.006 225 585 934 784 528 384 × 2 = 0 + 0.012 451 171 869 569 056 768;
  • 25) 0.012 451 171 869 569 056 768 × 2 = 0 + 0.024 902 343 739 138 113 536;
  • 26) 0.024 902 343 739 138 113 536 × 2 = 0 + 0.049 804 687 478 276 227 072;
  • 27) 0.049 804 687 478 276 227 072 × 2 = 0 + 0.099 609 374 956 552 454 144;
  • 28) 0.099 609 374 956 552 454 144 × 2 = 0 + 0.199 218 749 913 104 908 288;
  • 29) 0.199 218 749 913 104 908 288 × 2 = 0 + 0.398 437 499 826 209 816 576;
  • 30) 0.398 437 499 826 209 816 576 × 2 = 0 + 0.796 874 999 652 419 633 152;
  • 31) 0.796 874 999 652 419 633 152 × 2 = 1 + 0.593 749 999 304 839 266 304;
  • 32) 0.593 749 999 304 839 266 304 × 2 = 1 + 0.187 499 998 609 678 532 608;
  • 33) 0.187 499 998 609 678 532 608 × 2 = 0 + 0.374 999 997 219 357 065 216;
  • 34) 0.374 999 997 219 357 065 216 × 2 = 0 + 0.749 999 994 438 714 130 432;
  • 35) 0.749 999 994 438 714 130 432 × 2 = 1 + 0.499 999 988 877 428 260 864;
  • 36) 0.499 999 988 877 428 260 864 × 2 = 0 + 0.999 999 977 754 856 521 728;
  • 37) 0.999 999 977 754 856 521 728 × 2 = 1 + 0.999 999 955 509 713 043 456;
  • 38) 0.999 999 955 509 713 043 456 × 2 = 1 + 0.999 999 911 019 426 086 912;
  • 39) 0.999 999 911 019 426 086 912 × 2 = 1 + 0.999 999 822 038 852 173 824;
  • 40) 0.999 999 822 038 852 173 824 × 2 = 1 + 0.999 999 644 077 704 347 648;
  • 41) 0.999 999 644 077 704 347 648 × 2 = 1 + 0.999 999 288 155 408 695 296;
  • 42) 0.999 999 288 155 408 695 296 × 2 = 1 + 0.999 998 576 310 817 390 592;
  • 43) 0.999 998 576 310 817 390 592 × 2 = 1 + 0.999 997 152 621 634 781 184;
  • 44) 0.999 997 152 621 634 781 184 × 2 = 1 + 0.999 994 305 243 269 562 368;
  • 45) 0.999 994 305 243 269 562 368 × 2 = 1 + 0.999 988 610 486 539 124 736;
  • 46) 0.999 988 610 486 539 124 736 × 2 = 1 + 0.999 977 220 973 078 249 472;
  • 47) 0.999 977 220 973 078 249 472 × 2 = 1 + 0.999 954 441 946 156 498 944;
  • 48) 0.999 954 441 946 156 498 944 × 2 = 1 + 0.999 908 883 892 312 997 888;
  • 49) 0.999 908 883 892 312 997 888 × 2 = 1 + 0.999 817 767 784 625 995 776;
  • 50) 0.999 817 767 784 625 995 776 × 2 = 1 + 0.999 635 535 569 251 991 552;
  • 51) 0.999 635 535 569 251 991 552 × 2 = 1 + 0.999 271 071 138 503 983 104;
  • 52) 0.999 271 071 138 503 983 104 × 2 = 1 + 0.998 542 142 277 007 966 208;
  • 53) 0.998 542 142 277 007 966 208 × 2 = 1 + 0.997 084 284 554 015 932 416;
  • 54) 0.997 084 284 554 015 932 416 × 2 = 1 + 0.994 168 569 108 031 864 832;

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:


0.000 000 000 742 147 676 323(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 323(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:


0.000 000 000 742 147 676 323(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 -0.000 000 000 742 147 676 323 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 355 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