-0.000 000 000 742 147 676 473 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 473(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
-0.000 000 000 742 147 676 473(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:

|-0.000 000 000 742 147 676 473| = 0.000 000 000 742 147 676 473


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: 0.000 000 000 742 147 676 473.

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) 0.000 000 000 742 147 676 473 × 2 = 0 + 0.000 000 001 484 295 352 946;
  • 2) 0.000 000 001 484 295 352 946 × 2 = 0 + 0.000 000 002 968 590 705 892;
  • 3) 0.000 000 002 968 590 705 892 × 2 = 0 + 0.000 000 005 937 181 411 784;
  • 4) 0.000 000 005 937 181 411 784 × 2 = 0 + 0.000 000 011 874 362 823 568;
  • 5) 0.000 000 011 874 362 823 568 × 2 = 0 + 0.000 000 023 748 725 647 136;
  • 6) 0.000 000 023 748 725 647 136 × 2 = 0 + 0.000 000 047 497 451 294 272;
  • 7) 0.000 000 047 497 451 294 272 × 2 = 0 + 0.000 000 094 994 902 588 544;
  • 8) 0.000 000 094 994 902 588 544 × 2 = 0 + 0.000 000 189 989 805 177 088;
  • 9) 0.000 000 189 989 805 177 088 × 2 = 0 + 0.000 000 379 979 610 354 176;
  • 10) 0.000 000 379 979 610 354 176 × 2 = 0 + 0.000 000 759 959 220 708 352;
  • 11) 0.000 000 759 959 220 708 352 × 2 = 0 + 0.000 001 519 918 441 416 704;
  • 12) 0.000 001 519 918 441 416 704 × 2 = 0 + 0.000 003 039 836 882 833 408;
  • 13) 0.000 003 039 836 882 833 408 × 2 = 0 + 0.000 006 079 673 765 666 816;
  • 14) 0.000 006 079 673 765 666 816 × 2 = 0 + 0.000 012 159 347 531 333 632;
  • 15) 0.000 012 159 347 531 333 632 × 2 = 0 + 0.000 024 318 695 062 667 264;
  • 16) 0.000 024 318 695 062 667 264 × 2 = 0 + 0.000 048 637 390 125 334 528;
  • 17) 0.000 048 637 390 125 334 528 × 2 = 0 + 0.000 097 274 780 250 669 056;
  • 18) 0.000 097 274 780 250 669 056 × 2 = 0 + 0.000 194 549 560 501 338 112;
  • 19) 0.000 194 549 560 501 338 112 × 2 = 0 + 0.000 389 099 121 002 676 224;
  • 20) 0.000 389 099 121 002 676 224 × 2 = 0 + 0.000 778 198 242 005 352 448;
  • 21) 0.000 778 198 242 005 352 448 × 2 = 0 + 0.001 556 396 484 010 704 896;
  • 22) 0.001 556 396 484 010 704 896 × 2 = 0 + 0.003 112 792 968 021 409 792;
  • 23) 0.003 112 792 968 021 409 792 × 2 = 0 + 0.006 225 585 936 042 819 584;
  • 24) 0.006 225 585 936 042 819 584 × 2 = 0 + 0.012 451 171 872 085 639 168;
  • 25) 0.012 451 171 872 085 639 168 × 2 = 0 + 0.024 902 343 744 171 278 336;
  • 26) 0.024 902 343 744 171 278 336 × 2 = 0 + 0.049 804 687 488 342 556 672;
  • 27) 0.049 804 687 488 342 556 672 × 2 = 0 + 0.099 609 374 976 685 113 344;
  • 28) 0.099 609 374 976 685 113 344 × 2 = 0 + 0.199 218 749 953 370 226 688;
  • 29) 0.199 218 749 953 370 226 688 × 2 = 0 + 0.398 437 499 906 740 453 376;
  • 30) 0.398 437 499 906 740 453 376 × 2 = 0 + 0.796 874 999 813 480 906 752;
  • 31) 0.796 874 999 813 480 906 752 × 2 = 1 + 0.593 749 999 626 961 813 504;
  • 32) 0.593 749 999 626 961 813 504 × 2 = 1 + 0.187 499 999 253 923 627 008;
  • 33) 0.187 499 999 253 923 627 008 × 2 = 0 + 0.374 999 998 507 847 254 016;
  • 34) 0.374 999 998 507 847 254 016 × 2 = 0 + 0.749 999 997 015 694 508 032;
  • 35) 0.749 999 997 015 694 508 032 × 2 = 1 + 0.499 999 994 031 389 016 064;
  • 36) 0.499 999 994 031 389 016 064 × 2 = 0 + 0.999 999 988 062 778 032 128;
  • 37) 0.999 999 988 062 778 032 128 × 2 = 1 + 0.999 999 976 125 556 064 256;
  • 38) 0.999 999 976 125 556 064 256 × 2 = 1 + 0.999 999 952 251 112 128 512;
  • 39) 0.999 999 952 251 112 128 512 × 2 = 1 + 0.999 999 904 502 224 257 024;
  • 40) 0.999 999 904 502 224 257 024 × 2 = 1 + 0.999 999 809 004 448 514 048;
  • 41) 0.999 999 809 004 448 514 048 × 2 = 1 + 0.999 999 618 008 897 028 096;
  • 42) 0.999 999 618 008 897 028 096 × 2 = 1 + 0.999 999 236 017 794 056 192;
  • 43) 0.999 999 236 017 794 056 192 × 2 = 1 + 0.999 998 472 035 588 112 384;
  • 44) 0.999 998 472 035 588 112 384 × 2 = 1 + 0.999 996 944 071 176 224 768;
  • 45) 0.999 996 944 071 176 224 768 × 2 = 1 + 0.999 993 888 142 352 449 536;
  • 46) 0.999 993 888 142 352 449 536 × 2 = 1 + 0.999 987 776 284 704 899 072;
  • 47) 0.999 987 776 284 704 899 072 × 2 = 1 + 0.999 975 552 569 409 798 144;
  • 48) 0.999 975 552 569 409 798 144 × 2 = 1 + 0.999 951 105 138 819 596 288;
  • 49) 0.999 951 105 138 819 596 288 × 2 = 1 + 0.999 902 210 277 639 192 576;
  • 50) 0.999 902 210 277 639 192 576 × 2 = 1 + 0.999 804 420 555 278 385 152;
  • 51) 0.999 804 420 555 278 385 152 × 2 = 1 + 0.999 608 841 110 556 770 304;
  • 52) 0.999 608 841 110 556 770 304 × 2 = 1 + 0.999 217 682 221 113 540 608;
  • 53) 0.999 217 682 221 113 540 608 × 2 = 1 + 0.998 435 364 442 227 081 216;
  • 54) 0.998 435 364 442 227 081 216 × 2 = 1 + 0.996 870 728 884 454 162 432;

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:


0.000 000 000 742 147 676 473(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 473(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:


0.000 000 000 742 147 676 473(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 -0.000 000 000 742 147 676 473 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 425 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