1.401 298 464 324 817 070 923 729 583 289 916 131 271 35 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.401 298 464 324 817 070 923 729 583 289 916 131 271 35(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
1.401 298 464 324 817 070 923 729 583 289 916 131 271 35(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.401 298 464 324 817 070 923 729 583 289 916 131 271 35.

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) 0.401 298 464 324 817 070 923 729 583 289 916 131 271 35 × 2 = 0 + 0.802 596 928 649 634 141 847 459 166 579 832 262 542 7;
  • 2) 0.802 596 928 649 634 141 847 459 166 579 832 262 542 7 × 2 = 1 + 0.605 193 857 299 268 283 694 918 333 159 664 525 085 4;
  • 3) 0.605 193 857 299 268 283 694 918 333 159 664 525 085 4 × 2 = 1 + 0.210 387 714 598 536 567 389 836 666 319 329 050 170 8;
  • 4) 0.210 387 714 598 536 567 389 836 666 319 329 050 170 8 × 2 = 0 + 0.420 775 429 197 073 134 779 673 332 638 658 100 341 6;
  • 5) 0.420 775 429 197 073 134 779 673 332 638 658 100 341 6 × 2 = 0 + 0.841 550 858 394 146 269 559 346 665 277 316 200 683 2;
  • 6) 0.841 550 858 394 146 269 559 346 665 277 316 200 683 2 × 2 = 1 + 0.683 101 716 788 292 539 118 693 330 554 632 401 366 4;
  • 7) 0.683 101 716 788 292 539 118 693 330 554 632 401 366 4 × 2 = 1 + 0.366 203 433 576 585 078 237 386 661 109 264 802 732 8;
  • 8) 0.366 203 433 576 585 078 237 386 661 109 264 802 732 8 × 2 = 0 + 0.732 406 867 153 170 156 474 773 322 218 529 605 465 6;
  • 9) 0.732 406 867 153 170 156 474 773 322 218 529 605 465 6 × 2 = 1 + 0.464 813 734 306 340 312 949 546 644 437 059 210 931 2;
  • 10) 0.464 813 734 306 340 312 949 546 644 437 059 210 931 2 × 2 = 0 + 0.929 627 468 612 680 625 899 093 288 874 118 421 862 4;
  • 11) 0.929 627 468 612 680 625 899 093 288 874 118 421 862 4 × 2 = 1 + 0.859 254 937 225 361 251 798 186 577 748 236 843 724 8;
  • 12) 0.859 254 937 225 361 251 798 186 577 748 236 843 724 8 × 2 = 1 + 0.718 509 874 450 722 503 596 373 155 496 473 687 449 6;
  • 13) 0.718 509 874 450 722 503 596 373 155 496 473 687 449 6 × 2 = 1 + 0.437 019 748 901 445 007 192 746 310 992 947 374 899 2;
  • 14) 0.437 019 748 901 445 007 192 746 310 992 947 374 899 2 × 2 = 0 + 0.874 039 497 802 890 014 385 492 621 985 894 749 798 4;
  • 15) 0.874 039 497 802 890 014 385 492 621 985 894 749 798 4 × 2 = 1 + 0.748 078 995 605 780 028 770 985 243 971 789 499 596 8;
  • 16) 0.748 078 995 605 780 028 770 985 243 971 789 499 596 8 × 2 = 1 + 0.496 157 991 211 560 057 541 970 487 943 578 999 193 6;
  • 17) 0.496 157 991 211 560 057 541 970 487 943 578 999 193 6 × 2 = 0 + 0.992 315 982 423 120 115 083 940 975 887 157 998 387 2;
  • 18) 0.992 315 982 423 120 115 083 940 975 887 157 998 387 2 × 2 = 1 + 0.984 631 964 846 240 230 167 881 951 774 315 996 774 4;
  • 19) 0.984 631 964 846 240 230 167 881 951 774 315 996 774 4 × 2 = 1 + 0.969 263 929 692 480 460 335 763 903 548 631 993 548 8;
  • 20) 0.969 263 929 692 480 460 335 763 903 548 631 993 548 8 × 2 = 1 + 0.938 527 859 384 960 920 671 527 807 097 263 987 097 6;
  • 21) 0.938 527 859 384 960 920 671 527 807 097 263 987 097 6 × 2 = 1 + 0.877 055 718 769 921 841 343 055 614 194 527 974 195 2;
  • 22) 0.877 055 718 769 921 841 343 055 614 194 527 974 195 2 × 2 = 1 + 0.754 111 437 539 843 682 686 111 228 389 055 948 390 4;
  • 23) 0.754 111 437 539 843 682 686 111 228 389 055 948 390 4 × 2 = 1 + 0.508 222 875 079 687 365 372 222 456 778 111 896 780 8;
  • 24) 0.508 222 875 079 687 365 372 222 456 778 111 896 780 8 × 2 = 1 + 0.016 445 750 159 374 730 744 444 913 556 223 793 561 6;

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).


4. 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:


0.401 298 464 324 817 070 923 729 583 289 916 131 271 35(10) =

0.0110 0110 1011 1011 0111 1111(2)

5. Positive number before normalization:

1.401 298 464 324 817 070 923 729 583 289 916 131 271 35(10) =

1.0110 0110 1011 1011 0111 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.401 298 464 324 817 070 923 729 583 289 916 131 271 35(10) =

1.0110 0110 1011 1011 0111 1111(2) =

1.0110 0110 1011 1011 0111 1111(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0110 0110 1011 1011 0111 1111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


9. 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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

127(10) =

0111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 011 0011 0101 1101 1011 1111 1 =

011 0011 0101 1101 1011 1111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
011 0011 0101 1101 1011 1111


Decimal number 1.401 298 464 324 817 070 923 729 583 289 916 131 271 35 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 011 0011 0101 1101 1011 1111

Share

» Convert decimal 1.401 298 464 324 817 070 923 729 583 289 916 131 272 06 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 07, 2026 [July]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 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