-0.000 000 000 742 147 676 094 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 094(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
-0.000 000 000 742 147 676 094(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:

|-0.000 000 000 742 147 676 094| = 0.000 000 000 742 147 676 094


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: 0.000 000 000 742 147 676 094.

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) 0.000 000 000 742 147 676 094 × 2 = 0 + 0.000 000 001 484 295 352 188;
  • 2) 0.000 000 001 484 295 352 188 × 2 = 0 + 0.000 000 002 968 590 704 376;
  • 3) 0.000 000 002 968 590 704 376 × 2 = 0 + 0.000 000 005 937 181 408 752;
  • 4) 0.000 000 005 937 181 408 752 × 2 = 0 + 0.000 000 011 874 362 817 504;
  • 5) 0.000 000 011 874 362 817 504 × 2 = 0 + 0.000 000 023 748 725 635 008;
  • 6) 0.000 000 023 748 725 635 008 × 2 = 0 + 0.000 000 047 497 451 270 016;
  • 7) 0.000 000 047 497 451 270 016 × 2 = 0 + 0.000 000 094 994 902 540 032;
  • 8) 0.000 000 094 994 902 540 032 × 2 = 0 + 0.000 000 189 989 805 080 064;
  • 9) 0.000 000 189 989 805 080 064 × 2 = 0 + 0.000 000 379 979 610 160 128;
  • 10) 0.000 000 379 979 610 160 128 × 2 = 0 + 0.000 000 759 959 220 320 256;
  • 11) 0.000 000 759 959 220 320 256 × 2 = 0 + 0.000 001 519 918 440 640 512;
  • 12) 0.000 001 519 918 440 640 512 × 2 = 0 + 0.000 003 039 836 881 281 024;
  • 13) 0.000 003 039 836 881 281 024 × 2 = 0 + 0.000 006 079 673 762 562 048;
  • 14) 0.000 006 079 673 762 562 048 × 2 = 0 + 0.000 012 159 347 525 124 096;
  • 15) 0.000 012 159 347 525 124 096 × 2 = 0 + 0.000 024 318 695 050 248 192;
  • 16) 0.000 024 318 695 050 248 192 × 2 = 0 + 0.000 048 637 390 100 496 384;
  • 17) 0.000 048 637 390 100 496 384 × 2 = 0 + 0.000 097 274 780 200 992 768;
  • 18) 0.000 097 274 780 200 992 768 × 2 = 0 + 0.000 194 549 560 401 985 536;
  • 19) 0.000 194 549 560 401 985 536 × 2 = 0 + 0.000 389 099 120 803 971 072;
  • 20) 0.000 389 099 120 803 971 072 × 2 = 0 + 0.000 778 198 241 607 942 144;
  • 21) 0.000 778 198 241 607 942 144 × 2 = 0 + 0.001 556 396 483 215 884 288;
  • 22) 0.001 556 396 483 215 884 288 × 2 = 0 + 0.003 112 792 966 431 768 576;
  • 23) 0.003 112 792 966 431 768 576 × 2 = 0 + 0.006 225 585 932 863 537 152;
  • 24) 0.006 225 585 932 863 537 152 × 2 = 0 + 0.012 451 171 865 727 074 304;
  • 25) 0.012 451 171 865 727 074 304 × 2 = 0 + 0.024 902 343 731 454 148 608;
  • 26) 0.024 902 343 731 454 148 608 × 2 = 0 + 0.049 804 687 462 908 297 216;
  • 27) 0.049 804 687 462 908 297 216 × 2 = 0 + 0.099 609 374 925 816 594 432;
  • 28) 0.099 609 374 925 816 594 432 × 2 = 0 + 0.199 218 749 851 633 188 864;
  • 29) 0.199 218 749 851 633 188 864 × 2 = 0 + 0.398 437 499 703 266 377 728;
  • 30) 0.398 437 499 703 266 377 728 × 2 = 0 + 0.796 874 999 406 532 755 456;
  • 31) 0.796 874 999 406 532 755 456 × 2 = 1 + 0.593 749 998 813 065 510 912;
  • 32) 0.593 749 998 813 065 510 912 × 2 = 1 + 0.187 499 997 626 131 021 824;
  • 33) 0.187 499 997 626 131 021 824 × 2 = 0 + 0.374 999 995 252 262 043 648;
  • 34) 0.374 999 995 252 262 043 648 × 2 = 0 + 0.749 999 990 504 524 087 296;
  • 35) 0.749 999 990 504 524 087 296 × 2 = 1 + 0.499 999 981 009 048 174 592;
  • 36) 0.499 999 981 009 048 174 592 × 2 = 0 + 0.999 999 962 018 096 349 184;
  • 37) 0.999 999 962 018 096 349 184 × 2 = 1 + 0.999 999 924 036 192 698 368;
  • 38) 0.999 999 924 036 192 698 368 × 2 = 1 + 0.999 999 848 072 385 396 736;
  • 39) 0.999 999 848 072 385 396 736 × 2 = 1 + 0.999 999 696 144 770 793 472;
  • 40) 0.999 999 696 144 770 793 472 × 2 = 1 + 0.999 999 392 289 541 586 944;
  • 41) 0.999 999 392 289 541 586 944 × 2 = 1 + 0.999 998 784 579 083 173 888;
  • 42) 0.999 998 784 579 083 173 888 × 2 = 1 + 0.999 997 569 158 166 347 776;
  • 43) 0.999 997 569 158 166 347 776 × 2 = 1 + 0.999 995 138 316 332 695 552;
  • 44) 0.999 995 138 316 332 695 552 × 2 = 1 + 0.999 990 276 632 665 391 104;
  • 45) 0.999 990 276 632 665 391 104 × 2 = 1 + 0.999 980 553 265 330 782 208;
  • 46) 0.999 980 553 265 330 782 208 × 2 = 1 + 0.999 961 106 530 661 564 416;
  • 47) 0.999 961 106 530 661 564 416 × 2 = 1 + 0.999 922 213 061 323 128 832;
  • 48) 0.999 922 213 061 323 128 832 × 2 = 1 + 0.999 844 426 122 646 257 664;
  • 49) 0.999 844 426 122 646 257 664 × 2 = 1 + 0.999 688 852 245 292 515 328;
  • 50) 0.999 688 852 245 292 515 328 × 2 = 1 + 0.999 377 704 490 585 030 656;
  • 51) 0.999 377 704 490 585 030 656 × 2 = 1 + 0.998 755 408 981 170 061 312;
  • 52) 0.998 755 408 981 170 061 312 × 2 = 1 + 0.997 510 817 962 340 122 624;
  • 53) 0.997 510 817 962 340 122 624 × 2 = 1 + 0.995 021 635 924 680 245 248;
  • 54) 0.995 021 635 924 680 245 248 × 2 = 1 + 0.990 043 271 849 360 490 496;

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:


0.000 000 000 742 147 676 094(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 094(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:


0.000 000 000 742 147 676 094(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 -0.000 000 000 742 147 676 094 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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