-0.000 000 000 742 147 676 646 727 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 727 7(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
-0.000 000 000 742 147 676 646 727 7(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:

|-0.000 000 000 742 147 676 646 727 7| = 0.000 000 000 742 147 676 646 727 7


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: 0.000 000 000 742 147 676 646 727 7.

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) 0.000 000 000 742 147 676 646 727 7 × 2 = 0 + 0.000 000 001 484 295 353 293 455 4;
  • 2) 0.000 000 001 484 295 353 293 455 4 × 2 = 0 + 0.000 000 002 968 590 706 586 910 8;
  • 3) 0.000 000 002 968 590 706 586 910 8 × 2 = 0 + 0.000 000 005 937 181 413 173 821 6;
  • 4) 0.000 000 005 937 181 413 173 821 6 × 2 = 0 + 0.000 000 011 874 362 826 347 643 2;
  • 5) 0.000 000 011 874 362 826 347 643 2 × 2 = 0 + 0.000 000 023 748 725 652 695 286 4;
  • 6) 0.000 000 023 748 725 652 695 286 4 × 2 = 0 + 0.000 000 047 497 451 305 390 572 8;
  • 7) 0.000 000 047 497 451 305 390 572 8 × 2 = 0 + 0.000 000 094 994 902 610 781 145 6;
  • 8) 0.000 000 094 994 902 610 781 145 6 × 2 = 0 + 0.000 000 189 989 805 221 562 291 2;
  • 9) 0.000 000 189 989 805 221 562 291 2 × 2 = 0 + 0.000 000 379 979 610 443 124 582 4;
  • 10) 0.000 000 379 979 610 443 124 582 4 × 2 = 0 + 0.000 000 759 959 220 886 249 164 8;
  • 11) 0.000 000 759 959 220 886 249 164 8 × 2 = 0 + 0.000 001 519 918 441 772 498 329 6;
  • 12) 0.000 001 519 918 441 772 498 329 6 × 2 = 0 + 0.000 003 039 836 883 544 996 659 2;
  • 13) 0.000 003 039 836 883 544 996 659 2 × 2 = 0 + 0.000 006 079 673 767 089 993 318 4;
  • 14) 0.000 006 079 673 767 089 993 318 4 × 2 = 0 + 0.000 012 159 347 534 179 986 636 8;
  • 15) 0.000 012 159 347 534 179 986 636 8 × 2 = 0 + 0.000 024 318 695 068 359 973 273 6;
  • 16) 0.000 024 318 695 068 359 973 273 6 × 2 = 0 + 0.000 048 637 390 136 719 946 547 2;
  • 17) 0.000 048 637 390 136 719 946 547 2 × 2 = 0 + 0.000 097 274 780 273 439 893 094 4;
  • 18) 0.000 097 274 780 273 439 893 094 4 × 2 = 0 + 0.000 194 549 560 546 879 786 188 8;
  • 19) 0.000 194 549 560 546 879 786 188 8 × 2 = 0 + 0.000 389 099 121 093 759 572 377 6;
  • 20) 0.000 389 099 121 093 759 572 377 6 × 2 = 0 + 0.000 778 198 242 187 519 144 755 2;
  • 21) 0.000 778 198 242 187 519 144 755 2 × 2 = 0 + 0.001 556 396 484 375 038 289 510 4;
  • 22) 0.001 556 396 484 375 038 289 510 4 × 2 = 0 + 0.003 112 792 968 750 076 579 020 8;
  • 23) 0.003 112 792 968 750 076 579 020 8 × 2 = 0 + 0.006 225 585 937 500 153 158 041 6;
  • 24) 0.006 225 585 937 500 153 158 041 6 × 2 = 0 + 0.012 451 171 875 000 306 316 083 2;
  • 25) 0.012 451 171 875 000 306 316 083 2 × 2 = 0 + 0.024 902 343 750 000 612 632 166 4;
  • 26) 0.024 902 343 750 000 612 632 166 4 × 2 = 0 + 0.049 804 687 500 001 225 264 332 8;
  • 27) 0.049 804 687 500 001 225 264 332 8 × 2 = 0 + 0.099 609 375 000 002 450 528 665 6;
  • 28) 0.099 609 375 000 002 450 528 665 6 × 2 = 0 + 0.199 218 750 000 004 901 057 331 2;
  • 29) 0.199 218 750 000 004 901 057 331 2 × 2 = 0 + 0.398 437 500 000 009 802 114 662 4;
  • 30) 0.398 437 500 000 009 802 114 662 4 × 2 = 0 + 0.796 875 000 000 019 604 229 324 8;
  • 31) 0.796 875 000 000 019 604 229 324 8 × 2 = 1 + 0.593 750 000 000 039 208 458 649 6;
  • 32) 0.593 750 000 000 039 208 458 649 6 × 2 = 1 + 0.187 500 000 000 078 416 917 299 2;
  • 33) 0.187 500 000 000 078 416 917 299 2 × 2 = 0 + 0.375 000 000 000 156 833 834 598 4;
  • 34) 0.375 000 000 000 156 833 834 598 4 × 2 = 0 + 0.750 000 000 000 313 667 669 196 8;
  • 35) 0.750 000 000 000 313 667 669 196 8 × 2 = 1 + 0.500 000 000 000 627 335 338 393 6;
  • 36) 0.500 000 000 000 627 335 338 393 6 × 2 = 1 + 0.000 000 000 001 254 670 676 787 2;
  • 37) 0.000 000 000 001 254 670 676 787 2 × 2 = 0 + 0.000 000 000 002 509 341 353 574 4;
  • 38) 0.000 000 000 002 509 341 353 574 4 × 2 = 0 + 0.000 000 000 005 018 682 707 148 8;
  • 39) 0.000 000 000 005 018 682 707 148 8 × 2 = 0 + 0.000 000 000 010 037 365 414 297 6;
  • 40) 0.000 000 000 010 037 365 414 297 6 × 2 = 0 + 0.000 000 000 020 074 730 828 595 2;
  • 41) 0.000 000 000 020 074 730 828 595 2 × 2 = 0 + 0.000 000 000 040 149 461 657 190 4;
  • 42) 0.000 000 000 040 149 461 657 190 4 × 2 = 0 + 0.000 000 000 080 298 923 314 380 8;
  • 43) 0.000 000 000 080 298 923 314 380 8 × 2 = 0 + 0.000 000 000 160 597 846 628 761 6;
  • 44) 0.000 000 000 160 597 846 628 761 6 × 2 = 0 + 0.000 000 000 321 195 693 257 523 2;
  • 45) 0.000 000 000 321 195 693 257 523 2 × 2 = 0 + 0.000 000 000 642 391 386 515 046 4;
  • 46) 0.000 000 000 642 391 386 515 046 4 × 2 = 0 + 0.000 000 001 284 782 773 030 092 8;
  • 47) 0.000 000 001 284 782 773 030 092 8 × 2 = 0 + 0.000 000 002 569 565 546 060 185 6;
  • 48) 0.000 000 002 569 565 546 060 185 6 × 2 = 0 + 0.000 000 005 139 131 092 120 371 2;
  • 49) 0.000 000 005 139 131 092 120 371 2 × 2 = 0 + 0.000 000 010 278 262 184 240 742 4;
  • 50) 0.000 000 010 278 262 184 240 742 4 × 2 = 0 + 0.000 000 020 556 524 368 481 484 8;
  • 51) 0.000 000 020 556 524 368 481 484 8 × 2 = 0 + 0.000 000 041 113 048 736 962 969 6;
  • 52) 0.000 000 041 113 048 736 962 969 6 × 2 = 0 + 0.000 000 082 226 097 473 925 939 2;
  • 53) 0.000 000 082 226 097 473 925 939 2 × 2 = 0 + 0.000 000 164 452 194 947 851 878 4;
  • 54) 0.000 000 164 452 194 947 851 878 4 × 2 = 0 + 0.000 000 328 904 389 895 703 756 8;

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:


0.000 000 000 742 147 676 646 727 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 727 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 727 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 727 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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